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 $F/N\cong P$ be a presentation of $P$ with $F$ free of finite rank, and let $G$ be the torsion-free quotient $F/[N,N]$. Let $A$ denote the abelian normal subgroup $N/[N,N]$ of $G$.  Let $G^{(n)}$ denote the $n$th derived subgroup of $G$. Then $AG^{(n)}=G$ for all $n$ because $G/A$ is perfect. Since $G/A$ is finite and $A$ is abelian, there is an $n$ such that every solvable subgroup of $G$ has derived length at most $n$. For such an $n$, the subgroup $G^{(n)}$ is perfect and maps surjectively onto $P$. Moreover, $G$ and all its subgroups are closed flat manifold groups, being torsion-free, finitely generated, and abelian-by-finite.