In answer to the question "Is there a flat manifold with trivial first homology?" I proposed choosing a finite perfect group $P$ and a surjection $\phi:F\to P$ where $F$ is a free group of finite rank. If $K$ is the kernel of $\phi$ then the group $G:=[F,F]/[K,K]$ is perfect, torsion-free, finitely generated and abelian-by-finite.

This construction is not novel. I believe it is a standard strategy for obtaining torsion-free perfect groups.

The Hirsch length of $G$ is the dimension of a closed flat manifold with fundamental group $G$. One might hope to achieve the smallest possible Hirsch length by taking $P$ to be the alternating group $A_5$ and taking $F$ to be free on two generators. In that case the kernel $K$ is free on 61 generators by the Nielsen-Schreier Theorem, and $F/[K,K]$ has Hirsch length 61. Since $F/[F,F]$ has Hirsch length 2 we find that $G$, in this example, is a perfect f.g. abelian-by-finite group of Hirsch length 59. Is 59 the least dimension for which this phenomenon occurs? Maybe it is possible to factor out some of the rank 59 abelian normal subgroup in $G$ without sacrificing torsion-freeness. $G$, like any crystallographic group, has a unique largest abelian subgroup of finite index $A$ and it would be useful to know the constituents of the $\mathbb QP$-module $V:=A\otimes\mathbb Q$ as a $\mathbb QP$-module.

From this discussion, and the known fact that upto dimension 3 there are no perfect flat manifolds, the answer to the question lies in the range $4 \le n \le 59$. Can this be pinned down?