Here is an idea for making examples. Let $F$ be a free group and let $N$ be a normal subgroup of $F$. Then $F/[N,N]$ is torsion-free. To see this, suppose $w\in F$ has finite order modulo $[N,N]$, and let $E$ be the subgroup of $F$ generated by $w$ and $N$. Note that $E$ and $N$ are free groups, $E/[E,E]$ and $N/[N,N]$ are free abelian groups, and we have $[N,N]\subseteq[E,E]\subseteq N\subseteq E$. Therefore $E/[N,N]$ is torsion-free, being an extension of $[E,E]/[N,N]$ (which is a subgroup of $N/[N,N]$) by $E/[E,E]$. This being true for any choice of $w$ it follows that $F/[N,N]$ is torsion-free.
Now let $P$ be a finite perfect group. Let $F/N\cong P$ beChoose a presentation ofsurjective homomorphism $P$$F\twoheadrightarrow P$ with $F$ being a free group of finite rank,. Let $N$ be the kernel of this homomorphism and let $G$ bedenote the torsion-free quotient $F/[N,N]$. Let $A$ denote denote the abelian normal subgroup $N/[N,N]$ of $G$. Let $G^{(n)}$ denote theThen $n$th derived$A$ is a normal abelian subgroup of $G$ and $G/A$ is isomorphic to the finite perfect group $P$. ThenLet $AG^{(n)}=G$ for all$K$ be a normal subgroup of $n$ because$G$ such that $G/A$$G/K$ is perfectsolvable. SinceThen $G/A$$G/AK$ is finiteboth solvable and perfect and therefore $A$$AK=G$. It follows that $G/K = AK/K \cong A/A\cap K$ is abelian, there is an $n$ such that every. Thus the solvable subgroupquotients of $G$ has derived length at most $n$. For such an $n$,are abelian and the derived subgroup $G^{(n)}$$H:=[G,G]$ is perfect and maps surjectively onto $P$. Moreover,Since $G$ is finitely generated and abelian by finite, all its subgroups and quotients are closed flat manifold groups, being torsion-free, finitely generated, and abelian-by-finite by finite. Thus $H$ is abelian by finite and also normal in $G$. Since $HA=G$ we have $H/H\cap A$ is isomorphic to $P$. Now $H$ is a perfect closed flat manifold group mapping onto the original finite perfect group $P$.