Assuming that you DO mean that $T^n$ is a finite sheeted covering space, at the very least one can say that $\pi_1(M)$ is a torsion free $n$-dimensional crystallographic group.

If $n=3$, it follows that $M$ is homeomorphic to the closed Euclidean 3-manifold that one gets as the quotient of $\mathbb{E}^3$ by the 3-D crystallographic group isomorphic to $\pi_1(M)$; this follows from standard 3-manifold arguments. The complete list of these manifolds is known: amongst the 219 different 3-D crystallographic groups listed in http://en.wikipedia.org/wiki/Space_group, pick out the torsion free ones.

I don't know what happens beyond dimension 3, other than to say that $M$ is homotopy equivalent to the closed Euclidean $n$-manifold that one gets as the quotient of $\mathbb{E}^n$ by the $n$-dimensional crystallographic group isomorphic to $\pi_1(M)$.

Assuming that you DO mean that $T^n$ is a finite sheeted covering space, at the very least one can say that $\pi_1(M)$ is a torsion free $n$-dimensional crystallographic group. This follows from the Bieberbach theorems regarding groups that contain $\mathbb{Z}^n$ with finite index.

If $n=3$, it follows that $M$ is homeomorphic to the closed Euclidean 3-manifold that one gets as the quotient of $\mathbb{E}^3$ by the 3-D crystallographic group isomorphic to $\pi_1(M)$; this follows from standard 3-manifold arguments. The complete list of these manifolds is known: amongst the 219 different 3-D crystallographic groups listed in http://en.wikipedia.org/wiki/Space_group, pick out the torsion free ones.

I don't know what happens beyond dimension 3, other than to say that $M$ is homotopy equivalent to the closed Euclidean $n$-manifold that one gets as the quotient of $\mathbb{E}^n$ by the $n$-dimensional crystallographic group isomorphic to $\pi_1(M)$.