**Added.** Thanks to the comment of abx below I understood that there was a big gap in the reasoning, and $M$ doesn't need to be a fiber bundle.

**Example.** I'll construct an example $S$ of a complex surface that admits a submersion to an elliptic curve $E$ but that is not a fiber bundle over $E$. Then $M$ can be taken as $S\times E'$, where $E'$ is any elliptic curve (and $M$ admits a submersion to $E\times E'$).

So, let's start with the elliptic curve $E$ and any complex curve $C_1$ (say of genus $>2$) that admits a fixed point free involution $\sigma_1$. Now take any degree two cover $C_2$ of $E$ with ramifications (so that the genus of $C_2$ is $>1$). Let $\sigma_2$ be the corresponding involution of $C_2$. Finally consider the quotient of $C_1\times C_2$ by $\mathbb Z_2 $ that is acting on the $C_1$ factor by $\sigma_1$ and on the $C_2$ factor by $\sigma_2$. The resulting suface $(C_1\times C_2)/\mathbb Z_2$ admits a submersion to $E$.

**(corrected) Old answer.** The fact that such a manifold is a submersion implies that all fibers are smooth. As the above example shows, the submersion doesn't need to be a fiber bundle, but in case it *is* a fiber bundle something can be said.

Namely, one can say that the fiber bundle is isotrivial, i.e. all fibers are isomorphic curves. Indeed, we can associate to such a manifold a holomorphic map from $\mathbb C^2$ to the corresponding Teichmüller space, and since the latter space is a bounded domain, the map is constant.

At the same time we can't claim that the fibration is trivial, it can be isotrivial. Indeed, one can find curves $\Sigma$ that admit a non-trivial holomorphic action of $\mathbb Z^4$ on them. Then we can take a quotient of $\Sigma\times \mathbb C^2$ by such an action. It should not be hard to classify all such actions for small $2g-2=\chi(\Sigma)$, but I guess that for larger $g$ this might be difficult.