Consider $M^3_{pq}$, a torus bundle over $S^1$ with fundamental group the HNN extension generated by three generators $x,y,z$ satisfying the relations $\quad [x,y], \quad x^z = x^p \quad$ and $y^z = y^q$. By the exact homotopy sequence, $M^3_{pq}$ is aspherical.
Suppose there is a closed manifold $E^{n+3}$ that is the total space of a $T^n$ bundle over infinitely many $M^3_{pq}$ satisfying $p\neq q$. As $M_{pq}^3$ is aspherical, so is $E$. Hence, this can only be the case if $\pi_1 M_{pq}$ is a quotient of $\pi_1 E$ by the fundamental group $\pi_1T^n$; note that $\pi_1E$ and $n$ may not depend on $(p,q)$. It seems unlikely that one can obtain an infinite amount of mutually non-isomorphic, non-abelian groups as quotients from a finitely generated group by only identifying elements that commute. However, I have failed to prove this.
This begs the general question: How many exact sequences with non-abelian groups $H$
$$
0 \rightarrow Z^n \rightarrow G \rightarrow H \rightarrow e
$$
can a finitely presented group $G$ admit for fixed $n$? What are possible obstructions, aside from the kernel of $G\rightarrow H$ having to be abelian?