It is natural to compute $|SL_2(Z_{p^{e}})|$ via induction on $e$ as one has canonical exact sequence $SL_2(Z_{p^e})\to SL_2(Z_{p^{e-1}})\to 1$.
$\textbf{Q:}$ Recall that for $e=1$, $|SL_2(Z_p)|$'s computation is boiled down to computing linearly independent vectors as $Z_p$ is a field and this is easy. Is there an analogous way to compute this directly by combinatorial argument without going through induction? Say $v_1,v_2$ are linearly independent vectors of over $Z_{p^e}$ of $Z_{p^e}^2$. Now the question is $v_1\wedge v_2=det(v_1,v_2)$ necessarily invertible? It is not clear to me how I can guarantee invertiblity.
Yes, you can just count pairs of linearly independent vectors: $v_1$ and $v_2$ being linearly independent means that $a_1v_1 + a_2v_2 = 0$ implies $a_1 = a_2 = 0$. This means that the linear map $(a_1,a_2) \to a_1v_1 + a_2v_2$ is an injection from $(\mathbb{Z}/p^e)^2$ to itself and is therefore invertible. Moreover you can recover two linearly independent vectors from an invertible map by looking at where $(0,1)$ and $(1,0)$ go.