Setting $h=c=1$ the familiar Wikipedia result is that the Gravitational constant, $G$ goes like
$$ G = \frac{1}{m_{Planck}^2}$$ However, I suspect that this is only true in 4 or (3+1) dimensions? Or just 3 spatial dimensions. But how would this change if I work in d-dimensions? I guess I would have
$$ G_d = \frac{1}{m_{Planck}^{f(d)}}$$ where $f(d)$ is a dimensional dependent number. But then what is $f(d)$?
This is a fascinating subject!
I'll use "natural" units so that $\hbar = 1$ and $c = 1$. Poisson's equation in $d$ space dimensions is $$\tag{1} \nabla^2 \phi = 4 \pi G \rho, $$ where the mass density has units $\rho \sim \mathrm{L}^{-d - 1}$, since mass has units $m \sim \mathrm{L}^{-1}$ (in natural units). The left part of equation (1) has units $\nabla^2 \phi \sim \mathrm{L}^{-2}$, since $\phi \sim \mathrm{L}^0$ (no units! This comes from the potential energy or the force equation... and also from general relativity since $\phi$ is just a part of the spacetime metric). Thus it's easy to find the units of $G$: $$\tag{2} G \sim \mathrm{L}^{d - 1}. $$ For $d = 1$ (two spacetime dimensions), $G$ doesn't have any unit! For $d = 3$ space dimensions, we have $G \sim \mathrm{L}^{2}$ (the squared Planck lenght).
Notice that the electromagnetic fine-structure $\alpha \equiv \frac{k e^2}{\hbar c} \approx \frac{1}{137}$ have units for $d \ne 3$. The same exercice (using Maxwell equations) gives $$\tag{3} \alpha \sim \mathrm{L}^{d-3}. $$ It's interesting to note that the ratio $G/\alpha$ have units that don't depend on the number of dimensions: $$\tag{4} \frac{G}{\alpha} \sim \mathrm{L}^{2}. $$ Like gravitation (when $d \ne 1$), electromagnetism imposes a favored scale in universes of dimensions $d \ne 3$.
