Let $A$ denote the dimension of the line element $g_{μν} dx^μ dx^ν$, for the metric $g$, whose components with respect to the coordinates $x^μ$ are $g_{μν}$. Let $μ$ denote the dimension of the coordinate $x^μ$, and denote dimensions by brackets $[⋯]$. Then
$$\left\{\left[g_{μν} dx^μ dx^ν\right] = A,\quad \left[x^μ\right] = μ = \left[dx^μ\right]\right\}\quad→\quad \left[g_{μν}\right] = \frac{A}{μν}.$$
The usual convention is to adopt $A = L^2$, where $L$ denotes the dimension for length; and it is the convention that will be adopted here and below.
Dimensional-Analysis: as gauge fields
You want to line the dimensional analysis up with the corresponding dimensional analysis for gauge fields and to also tie everything together with $GL(4)$. We'll do that here ... and more.
For the following, let $∂_μ = ∂/∂x^μ$. Then
$$\left[∂_μ\right] = \frac1{μ}.$$
Dimensional-analysis: gravity as a $GL(4)$ gauge field
The dimension for the connection comes by ensuring that the terms in the expression for the covariant derivative of the metric
$$∇_μ g_{νρ} = ∂_μ g_{νρ} - Γ^σ_{μν} g_{σρ} - Γ^σ_{μρ} g_{νσ} \left(= 0\right).$$
should each have the same dimensions. For metrical connections, the covariant derivative is 0, but we won't need that anywhere here. In particular, from
$$\frac{A}{μνρ} = \left[∂_μ g_{νρ}\right] = \left[Γ^σ_{μν} g_{σρ}\right] = \left[Γ^σ_{μν}\right] \frac{A}{σρ}$$
follows
$$\left[Γ^σ_{μν}\right] = \frac{σ}{μν}.$$
This can be thought of as the components of a gauge field associated with $GL(4)$, written with respect to a Lie algebra basis $Y_a$ as $A_μ = A^a_μ Y_a$. Here, the basis index $a$ is represented as a double index $a = \left({}^ρ_ν\right)$, written as $Y_a = e^ν_ρ$, while the components $A^a_μ$ are written as $A^a_μ = Γ^ρ_{μν}$. The Lie bracket is given by
$$\left[e^μ_ν,e^ρ_σ\right] = δ^μ_σ e^ρ_ν - δ^ρ_ν e^μ_σ,$$
so the corresponding field strength
$$F_{μν} = ∂_μ A_ν - ∂_ν A_μ + \left[A_μ, A_ν\right],$$
is just $F_{μν} = F^a_{μν} Y_a$, which, for the index $a = \left({}^σ_ρ\right)$, yields
$$F^a_{μν} = R^σ_{ρμν} = ∂_μ Γ^σ_{νρ} - ∂_ν Γ^σ_{μρ} + Γ^σ_{μτ} Γ^τ_{νρ} - Γ^σ_{ντ} Γ^τ_{μρ}.$$
This is just the curvature tensor in component form, and its dimensions may be directly read off from this equation:
$$\left[R^σ_{ρμν}\right] = \frac{σ}{ρμν}.$$
The terms in the equation for $F_{μν}$ have matching dimensions with respect to each other provided the dimensions are set as:
$$\left[e^ρ_σ\right] = \frac{ρ}{σ},\quad \left[A_μ\right] = \frac1{μ},\quad \left[F_{μν}\right] = \frac1{μν}.$$
For the associated one-form $A = A_μ dx^μ$ and two-form $F = ½ F_{μν} dx^μ ∧ dx^ν$, the dimensions are just:
$$[A] = 1,\quad [F] = 1.$$
Dimensional-analysis: electromagnetism as a $U(1)$ gauge field
You can compare this to the dimensions for the electromagnetic field. Indexing the coordinates as $x^0 = t$, $\left(x^1, x^2, x^3\right)$ (in the flat-space Minkowski limit), the dimensions are $0 = T$, $1 = 2 = 3 = L$, where $T$ denotes the dimension of duration. Then the field potential $A_μ$ has components $A_0 = -φ$, the electric or scalar potential and $\left(A_1, A_2, A_3\right) = $, the magnetic or vector potential. Their dimensions are
$$[φ] = \frac{P}{T},\quad [] = \frac{P}{L},$$
where $P$ denotes the dimension of magnetic charge or magnetic flux. Correspondingly, the potential 1-form $A = A_μ dx^μ$ and its components $A_μ$ have dimensions
$$\left[A_μ dx^μ\right] = P,\quad \left[A_μ\right] = \frac{P}{μ}.$$
This can be reconciled with $[A] = 1$ by noting that the electromagnetic field - as a gauge field - is associated with a one-dimensional Lie algebra, so it has a hidden index $A_μ → A^γ_μ$ associated with the one-dimensional Lie algebra basis $\left(Y_γ\right)$. So, the actual dimensions for the potential 1-form and its components are:
$$\left[A^γ_μ\right] = \frac{P}{μ},\quad \left[A_μ\right] = \left[A^γ_μ Y_γ\right] = \frac{P}{μ} \left[Y_γ\right],\quad [A] = P \left[Y_γ\right],$$
and can be made consistent with the $[A] = 1$, by setting
$$\left[Y_γ\right] = \frac1{P}.$$
The resulting dimensions for the components $A_μ$ are, then,
$$\left[A_μ\right] = \frac1{μ}.$$
The corresponding field strength is given, component-wise, by $ = \left(F^γ_{23}, F^γ_{31}, F^γ_{12}\right)$ and $ = \left(F^γ_{10}, F^γ_{20}, F^γ_{30}\right)$, with the dimensions
$$[] = \frac{P}{L^2},\quad [] = \frac{P}{LT}.$$
This yields the resulting dimensions
$$\left[F^γ_{μν}\right] = \frac{P}{μν},\quad \left[F_{μν}\right] = \left[F^γ_{μν} Y_γ\right] = \frac{P}{μν} \frac1{P} = \frac1{μν},\quad [F] = 1.$$
So, that's how you reconcile the dimensional analysis for the gravity/inertia field with that for gauge fields, like the Maxwell field and align them all.
Extra commentary on the response fields, action and couplings
For the following, let $d^n x = dx^0 ∧ dx^1 ∧ ⋯ ∧ dx^{n-1}$ and $Ω = \left[d^n x\right]$. Then
$$Ω = 0 1 ⋯ (n-1).$$
Let $g^{μν}$ denote the components of the inverse metric, and $g$ the determinant of the metric. Then
$$g^{μν} = \frac{μν}{A},\quad g = \frac{A^n}{Ω^2}\quad →\quad \left[\sqrt{|g|} d^n x\right] = A^{n/2},$$
for $n$ dimensions independently of how you index the coordinates. Let $H$ denote the dimension of the action and of Planck's constant $h$. Then
$$[h] = H = \frac{ML^2}{T},$$
where $M$ is the dimension of mass. Finally, for an action integral $S = ∫ d^4x$ given by a Lagrangian density $$, the respective dimensions are
$$[S] = H,\quad [] = \frac{H}{Ω}.$$
Electro-magnetism
In the presence of an action $S = ∫ d^4x$, given by a Lagrangian density $$, the response fields would be given by
$$ = +\frac{∂}{∂},\quad = -\frac{∂}{∂},$$
the archetypical case being the Maxwell-Lorentz Lagrangian density
$$ = -\frac14 ε_0 c \sqrt{|g|} g^{μρ} g^{νσ} F_{μν} F_{ρσ} = ε_0 \frac{E^2 - B^2 c^2}2,\quad = ε_0 ,\quad = ε_0 c^2 .$$
The dimensions of the response fields are
$$[] = \frac{Q}{L^2},\quad [] = \frac{Q}{LT},$$
where $Q$ is the dimension for electric charge, which yields the following as the dimension for the Lagrangian density:
$$[] = \left[\frac{∂}{∂}\right] = \frac{[]}{[]}\quad⇒\quad [] = [][] = \frac{PQ}{L^3T}.$$
For the particular choice of coordinate indexing, this works out to:
$$Ω = \left[dt ∧ dx ∧ dy ∧ dz\right] = L^3 T,$$
thus
$$\frac{PQ}{L^3T} = [] = \frac{H}{L^3T}\quad⇒\quad PQ = H.$$
In SI, the units for $(P,Q,M,L,T)$ are, respectively, Weber, Coulomb, kilogram, meter and second, so Weber·Coulomb = kilogram·meter²/second - reflecting the widely-known fact that electric and magnetic charge are "complementary".
Gravity
For general relativity, the Einstein-Hilbert action $S = ∫ (1/2κ) R \sqrt{|g|} d^4 x$, where $κ$ is the associated coupling coefficient.
This involves the curvature scalar $R = g^{μν} R_{μν}$ and the components $R_{μν} = R^ρ_{νρμ}$ of the Ricci tensor, with the respective dimensions being:
$$\left[R_{μν}\right] = \frac1{μν},\quad \left[R\right] = \frac1{A}.$$
In $n = 4$ dimensions
$$\left[\sqrt{|g|} d^4x\right] = A^2.$$
Putting all of this together leads to the following for $[κ]$:
$$H = [S] = \left[\frac1{2κ}\right] [R] \left[\sqrt{|g|} d^4x\right] = \frac1{[κ]} \frac{1}{A} A^2\quad⇒\quad [κ] = \frac{A}{H} = \frac{T}{M}.$$
For a space-time of $n$ dimensions, this would generalize to $[κ_n] = A^{(n-2)/2}/H$.
The two wide-spread oopsies in the literature (and on the net)
Since the dimensions of Newton's coefficient and in-vacuuo light speed are, respectively
$$[G] = \frac{L^3}{MT^2},\quad [c] = \frac{L}{T},$$
then it follows that
$$[κ] = \frac{[G]}{[c]^3} = \left[\frac{8πG}{c^3}\right].$$
That is: the correct value of the coupling coefficient $κ$ is not $8πG/c^4$ - as stated nearly everywhere on the net - but
$$κ = \frac{8πG}{c^3}.$$
You'll notice, by the way, that they even try to fudge their way around that in the reference provided by the PDG listings by just arbitrarily sticking in an extra factor of $c$! From 21. Experimental Tests of Gravitational Theory
$$S_{\text{tot}}\left(g_{μν},ψ,A_μ,H\right) = c^{-1}\int d^4 x (_{\text{Ein}} + _{\text{SM}}),\tag{21.2}$$
$$_{\text{Ein}}\left(g_{αβ}\right) = \frac{c^4}{16πG} \int \sqrt{g}g^{μν} R_{μν}\left(g_{αβ}\right).\tag{21.3}$$
(And even that fix, of putting the $c^{-1}$ out the outside, botches it up, because now you have $_{\text{SM}}$ getting the wrong dimensions.)
In fact, one will see $8πG/c^n$ in the literature for $n = 2, 3, 4$. Einstein thought originally had it as $n = 2$. The wall over at Leiden University, with Einstein's equation on it, uses $n = 4$, but only $n = 3$ is correct.
Another wide-spread error is the use of $8π$ in the coefficient $κ_n$ for dimensions $n ≠ 4$, where (in fact) the number is specific to 4-dimensions and ultimately has to do with the surface area of the 3-sphere. The $n$-dimensional analogue $G_n$ of Newton's coefficient $G_4 = G$ would involve the area of the ($n-1$)-dimensional hyper-sphere and, after going through the analysis carefully (e.g. as in Gravitational coupling constant in higher dimensions), you'd end up getting $κ_n = A_n G_n/c^3$, where
$$A_n = \frac{(n-1)(n-2)}{n-3} \frac{π^{(n-1)/2}}{((n-1)/2)!},$$
adopting the convention that $(-½)! = \sqrt{π}$.
The dimension $[h κ_n] = A^{(n-2)/2}$, where $h$ is Planck's constant (with dimension $[h] = H$) will always that for the dimension of the $n - 2$ (hyper-)area in $n$ dimensions. That shows that in whatever way general relativity and quantum theory combine, there will probably be something that has to do with the quantization of (hyper-)areas. For 4-dimensions, the associated area is $hG$.
