Note that a branched covering induces an injection of rational cohomology rings, by transfer considerations. Therefore the cohomology of a manifold that is a branched covering of $T^3$ must contain three classes of degree 1 whose triple cup product is nontrivial.
This condition on a manifold $M^3$ implies the existence of a map $M^3\to T^3$ of nonzero degree. Passing to a covering space of $T^3$ if necessary we obtain such a map that is also surjective on $\pi_1$. Assuming the resulting map is of degree $\ge 3$, the main result of [Edmonds, Allan L.Deformation of maps to branched coverings in dimension three. Math. Ann. 245 (1979), no. 3, 273--279.] implies that this map is homotopic to a branched covering.