Q:

Proof of area or volume element

Why is the volume element equal to the square root of the absolute value of the metric tensor determinant, i.e. $$dV=\sqrt{\left | g \right |} dx^0dx^1dx^2$$

A:

First, the definitions. Let $e_1,\ldots,e_n$ be  an orthonormal basis, i.e. $\langle e_i,e_j\rangle=\delta_{ij}$, and $\theta_1,\ldots,\theta_n$ its dual, i.e. 
$\theta_i(e_j)=\delta_{ij}$.  The volume element is defined by $vol:=\theta_1\wedge\ldots\wedge\theta_n$. 
The rest is a simple calculation. If you change basis, say $e_i=\sum_j a_{ij} e'_j,$ with dual basis $\theta'_i$, then   $\theta_i=\sum_j b_{ij}\theta'_j,$ where the change of basis  matrices $A=(a_{ij}), B=(b_{ij})$, satisfy $AB^t=I$. Let $g_{ij}=\langle e'_i,e'_j\rangle.$ Then  the matrix $G=(g_{ij})$ satisfies $I=AGA^t$. 
Now plug-in $\theta_i=\sum_j b_{ij}\theta'_{j} $ into  $vol=\theta_1\wedge\ldots\wedge\theta_n$ and get $vol=\det(B)\theta'_1\wedge\ldots\wedge\theta'_n$. Taking determinants of the matrix equations $AB^t=AGA^t=I$, you get $g:=det(G)=det(B)^2.$ 
Note: there is a sign issue, having to do with orientation, but I guess this is not your concern here. 

