The first part of this answer is merely an elaboration on Marc Kegel and T. Amdeberhan's comments, and addresses the question before the edit. As they say, $T^3$ is prime by virtue of being irreducible (a stronger condition) -- since $\pi_2(T^3)$ is trivial, every embedded 2-sphere is null-homotopic in $T^3$ and it follows that it bounds a ball. [This wikipedia page](https://en.wikipedia.org/wiki/Prime_manifold) may be helpful for you.

While it is true that $S^1\times S^2$ is the only prime closed 3-manifold with fundamental group $\mathbb{Z}$ (and all infinite cyclic groups are isomorphic to $\mathbb{Z}$), this does not imply that $T^3$ is not prime. (On the off chance this helps, note that $\mathbb{Z}^3$ is not a cyclic group).

In the language of Hatcher's paper, $T^3$ is a prime 3-manifold of type III: it is $K(\mathbb{Z}^3,1)$.  Note also that $T^3$ is a Seifert manifold, thus the following statement in Hatcher's paper is consistent with this:

>The only Seifert manifold that is not prime is $\mathbb{RP}^3\#\mathbb{RP}^3$, the sum of two copies of real projective 3-space.

I believe the edit of your question is addressed in the answers to [this question](https://mathoverflow.net/questions/8223/reducible-3d-torus-bundles).  Namely, torus bundles are irreducible, so they admit no nontrivial connected sum decomposition.