Since you didn't specify, I'll assume this is in a Banach space $X$, with $T: X \to Y$ (for another Banach space $Y$) a closed densely-defined operator, and all $x_n \in \mathscr D(T)$.
By the uniform boundedness principle, any weakly convergent sequence is bounded. So a necessary condition is that $T x_n$ is bounded.
If $Y$ is reflexive and the dual operator $T': Y' \to X'$ also densely defined, that is also sufficient: $f(T x_n) = (T' f)(x_n)$ converges for $f \in \mathscr D(T')$, and the map $f \mapsto \lim_n f(T x_n)$ extends by continuity to a member of $Y''$ which corresponds to a member of $Y$, and then we verify that this is the weak limit of $T x_n$.