3D common way
Definition of angular momentum vector ($m = 1$)
\begin{equation}
L_l = \epsilon_{lij}x^iv^j
\end{equation}
($L_l$ dual to $L^{ij} = x^iv^j - x^jv^i$ tensor)
Rortational velocity of particle
\begin{equation}
v^j = \epsilon^{jrk}\omega_rx_k.
\end{equation}
($\omega_r$ dual to $\omega^{jk}$ antisymmetric tensor of angular velocity $ \epsilon^{jrk}\omega_r = \omega^{jk}$)
Substitute in the angular momentum definition
\begin{equation}
L_l = \epsilon_{lij}x^i\epsilon^{jrk}\omega_rx_k.
\end{equation}
Let us use the property of the Levi-Civita tensor:
\begin{equation}
\epsilon^{lij} = -\epsilon^{jli}
\end{equation}
then
\begin{equation}
L_l = - \epsilon_{jli}\epsilon^{jkr}x^ix_k\omega_r.
\end{equation}
Let us use another property of the Levi-Civita tensor:
\begin{equation}
\epsilon_{jli}\epsilon^{jkr} = \delta_l^k\delta_i^r - \delta_l^r\delta_i^k.
\end{equation}
\begin{equation}
L_l = \left( \delta_l^r\delta_i^k - \delta_l^k\delta_i^r \right) x^ix_k\omega_r.
\end{equation}
Expand the brackets and take into account that $\delta_i^r x^i = x^r$, $\delta_l^k x_k = x_l$ and $\delta_i^k x^i = x^k$, we get
\begin{equation}
L_l = \left( \delta_l^r x^kx_k - x^rx_l\right) \omega_r,
\end{equation}
or
\begin{equation}
L_l = \left( \delta_l^r x^2 - x^rx_l\right) \omega_r,
\end{equation}
where inertia tensor
\begin{equation}
I_l^r = \delta_l^r x^2 - x^rx_l
\end{equation}
Arbitrary number of dimensions
For Higher dimensions we can use directly angular momentum tensor
\begin{equation}
L^{ij} = x^iv^j - x^jv^i
\end{equation}
and
Rortational velocity of particle we can express via angular momentum tensor
\begin{equation}
v^j = \omega^{jm}x_m.
\end{equation}
Then
\begin{equation}
L^{ij} = x^ix_m\omega^{jm} - x^jx_m\omega^{im}
\end{equation}
Now we can lower indices near $\omega$ with metric tensor:
\begin{align}
\omega^{jm} = g^{jn}g^{mr}\omega_{nr} \\
\omega^{im} = g^{in}g^{mr}\omega_{nr}
\end{align}
So, we get
\begin{equation}
L^{ij} = \left( x^ix_mg^{jn}g^{mr} - x^jx_m g^{in}g^{mr}\right) \omega_{nr}
\end{equation}
So, we can conclude $L^{ij} = I^{ijnr} \omega_{nr}$, the inertia tensor is
\begin{equation}
I^{ijnr} = x^ix_mg^{jn}g^{mr} - x^jx_m g^{in}g^{mr}
\end{equation}
