yes! thak you, i think you're right, because given the resolution of singularities $Hilb^3(T)\rightarrow Sym^3(T)$ we have that generally the fiber on the singular locus is a $\mathbb{P}^1$, so intersecting with $Ker(\alpha)$ we can say $F$ is bimeromorphic to a $\mathbb{P}^1$-bundle on $T$, which of course has $T$ as Albanese variety!