This computation could in principle be done using Clifford theory. Clifford theory tells you how to describe the representation theory of a group $G$ given that it can be described as an extension
$$1 \to N \to G \to H \to 1$$
and $SL_d(\mathbb{Z}/p^n\mathbb{Z})$ can be described as an extension
$$1 \to N \to SL_d(\mathbb{Z}/p^n \mathbb{Z}) \to SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) \to 1$$
where $N$ is the kernel of the reduction $\bmod p^{n-1}$ map. It consists precisely of elements of $SL_d(\mathbb{Z}/p^n\mathbb{Z})$ congruent to $I \bmod p^{n-1}$, or equivalently of the form $I + p^{n-1} M$. Every such matrix is invertible over $\mathbb{Z}/p^n\mathbb{Z}$ and has determinant $1$ iff $\text{tr}(M) \equiv 0 \bmod p$, and multiplying two such matrices even shows that $N$ is isomorphic to the additive group of such matrices, hence
$$N \cong (\mathbb{Z}/p\mathbb{Z})^{d^2-1}.$$
If you like, you can think of $N$ as $\mathfrak{sl}_d(\mathbb{Z}/p\mathbb{Z})$.
Clifford theory simplifies substantially when $N$ is abelian, so this is helpful, although I think the detailed analysis will still be difficult to carry out, and doing it this way you'll have to repeat the analysis $n-1$ times to reduce to the case of $SL_d(\mathbb{Z}/p\mathbb{Z})$.