Let $M$ be a 3-manifold fibering over $S^1$, so there exists a fibration $\Sigma \to M \to S^1$. Then $M\times M$ will admit a symplectic structure.

There is a symplectic structure on $M\times S^1$, associated to the fibration $\Sigma \to M\times S^1 \to S^1\times S^1=T^2$ which is trivial in the second factor, by a result of Thurston (the converse also holds).

Similarly, there is a fibration $\Sigma \to M\times M \to S^1\times M$ which is trivial in the second factor. Hence by the result in the fourth paragraph of Thurston's paper, $M\times M$ admits a symplectic structure.

I think these manifolds cannot be Kähler usually. See the main result of Biswas-Mj-Seshadri.

Let $M$ be a 3-manifold fibering over $S^1$, so there exists a fibration $\Sigma \to M \to S^1$. Then $M\times M$ will admit a symplectic structure.

There is a symplectic structure on $M\times S^1$, associated to the fibration $\Sigma \to M\times S^1 \to S^1\times S^1=T^2$ which is trivial in the second factor, by a result of Thurston (the converse also holds).

Similarly, there is a fibration $\Sigma \to M\times M \to S^1\times M$ which is trivial in the second factor. Hence by the result in the fourth paragraph of Thurston's paper, $M\times M$ admits a symplectic structure.

These manifolds cannot be Kähler usually. See Theorem 1.2 of Biswas-Mj-Seshadri, which implies that if $M\times M$ is Kähler, then either $M$ is a manifold admitting Nil geometry (the fundamental group is commensurable with the Heisenberg group $H$), or $M$ is finitely covered by $\Sigma \times S^1$ for some surface $\Sigma$ of positive genus. See Dmitri Panov's answer for a non-trivial example. Question 4.4 in the paper leaves open the possibility of whether $H\times H$ can be Kähler.