Simple module over $ֿ\mathbb{Z}G$ has a $\mathbb{Z}N$ composition series when $N \triangleleft G$ is nilpotent and of finite-index

If $R$ is a ring, $G$ a group, and $N$ a normal subgroup of finite index, then any simple $RG$-module is semisimple and of finite length (at most $[G:N]$) as an $RN$-module. No other hypotheses about $G,N$ are needed. This result can be found in Passman's "The Algebraic Structure of Group Rings", Theorem 7.2.16. He cites a 1937 paper of A. H. Clifford.

Even more generally, if $A,B$ are rings such that $B$ is a finite normalizing extension of $A$, then any simple $B$-module is semisimple and of finite length as an $A$-module. This is Proposition 10.1.9(ii) in "Noncommutative Noetherian Rings" by McConnell and Robson -- see paragraph 10.1.5 for notation and assumptions. To say $B$ is a finite normalizing extension of $A$ means that there are elements $b_1,\dots,b_k$ that generate $B$ as a left (or right) $A$-module and satisfy $Ab_i=b_iA$ for each $i$.

Here is a sketch of the proof in the group ring case. Let $g_1,\dots,g_k$ be a complete set of coset representatives for $N$ and let $M$ be a simple left $RG$-module. It's easy to see $M$ is finitely generated as an $RN$-module, so it contains a maximal $RN$-submodule $L$. The intersection $\cap_{i=1}^k g_iL$ is an $RG$-submodule of $M$ and so must be zero. This implies that $M$ embeds as an $RN$-submodule in the direct sum of the $RN$-modules $M/g_iL$. Since $g_iL$ is a maximal $RN$-submodule of $M$, each of these factor modules is simple. Hence $M$ is semisimple as an $RN$-module, of length at most $k$.