It's not just torsion-free, we can actually compute it in terms of the Picard and Albanese variety of $T$, by the following classical result:
Lemma. If $X$ and $Y$ are smooth projective varieties over a field $k$, then there is a canonical short exact sequence of group schemes
$$0 \to \mathbf{Pic}_X \times \mathbf{Pic}_Y \to \mathbf{Pic}_{X \times Y} \to \mathbf{Hom}\big(\mathbf{Alb}^1_X,(\mathbf{Pic}^0_Y)^\text{red}\big) \to 0.$$
Here, $(\mathbf{Pic}^0_X)^\text{red}$ is an abelian variety; see for instance [FGA Explained, Thm. 9.5.4]. Its dual is denoted $\mathbf{Alb}^0_X$, and without choosing a base point this gives a torsor $\mathbf{Alb}^1_X$ under $\mathbf{Alb}^0_X$ together with a morphism $X \to \mathbf{Alb}^1_X$ that is initial in the category of maps from $X$ to torsors under abelian varieties. Finally, $\mathbf{Hom}$ denotes the (ind-étale) group scheme of morphisms of abelian varieties (or probably morphisms of pairs of an abelian variety with a torsor under it, up to translation ― once you choose a base point on $X$, this should be the same thing).
Torsion-freeness then follows since $\mathbf{Hom}$ schemes between abelian varieties are torsion-free.
For a proof of the lemma, see for instance my dissertation [vDdB, §4.4], although I claim no originality here! (That said, I also didn't find this precise statement in the literature...)
A different argument goes by choosing base points $x \in X(k)$ and $y \in Y(k)$ (first enlarge the field if necessary and worry about descent later). These give a section $\mathbf{Pic}_{X \times Y} \to \mathbf{Pic}_X \times \mathbf{Pic}_Y$. To compute the kernel, consider a line bundle $\mathscr L$ on $X \times Y \times T$ for some $k$-scheme $T$ that is trivial along $x \times Y \times T$ and $X \times y \times T$. The second trivialisation gives a morphism $X \times T \to \mathbf{Pic}_Y$ via the rigidified Picard functor, and the first says that the image lands in $\mathbf{Pic}_Y^0$. This gives a morphism of $T$-schemes $X \times T \to \mathbf{Pic}_Y^0 \times T$, which by the universal property of $\mathbf{Alb}^1_X$ corresponds to a $T$-point of $\mathbf{Hom}(\mathbf{Alb}^1_X,(\mathbf{Pic}^0_Y)^\text{red})$. Now check that this doesn't depend on the choice of section and hence descends down to $k$.
References.
[FGA Explained] B. Fantechi, L. Göttsche, L. Illusie, S. L. Kleiman, N. Nitsure, A. Vistoli, Fundamental algebraic geometry: Grothendieck’s FGA explained. Mathematical Surveys and Monographs 123. American Mathematical Society, 2005. ZBL1085.14001.
For Kleiman's chapter, see also arXiv:math/0504020
[vDdB] R. van Dobben de Bruyn, Dominating varieties by liftables ones. PhD thesis, Columbia University, 2018.