In the generality stated, the question has a negative answer (which took me embarrassingly long to spot); the point is that when $T_1$ and $T_3$ are not assumed injective or surjective they give us very little traction on what $T_2$ is, in contrast to the intuition one might have from the Five Lemma. The "silly" counterexample might be useful for other settings so here are the details; it should work in any category with zero object and an appropriate notion of "binary sum".

Take objects (=Banach spaces if you wish to be concrete) $A$ and $B$ and let $\iota_L: A\to A\oplus B$, $\pi_L: A\oplus B \to A$, $\iota_R:B\to A\oplus B$ and $\pi_R: A\oplus B \to B$ be the usual embedding and projection operators.

Take as your top row the short exact sequence $A \stackrel{\iota_L}{\to} A\oplus B \stackrel{\pi_R}{\to} B$ and as your bottom row the short exact sequence $B \stackrel{\iota_R}{\to} A\oplus B \stackrel{\pi_L}{\to} A$. The vertical arrow on the left is the zero map $A\to B$, the vertical arrow on the right is the zero map $B\to A$, and the middle arrow is $(0,{\rm id_B}) : A\oplus B \to A\oplus B$. Then everything commutes.

To turn this into a counterexample for the original question, just take $B$ to be your favourite infinite-dimensional Banach space and $A$ to be an arbitrary Banach space.

On the other hand, I think I can prove the following: suppose I am given everything in your initial diagram except the middle arrow $T_2$, so that we have (strict) exactness on the top row and bottom row and nuclear operators $T_1:X_1\to Y_1$, $T_3:X_3\to Y_3$. Then $T_1$ has a nuclear extension $R:X_2\to Y_1$, $T_3$ has a nuclear lift $S:X_3\to Y_2$, and defining $\theta= g_1R+Sf_2$ gives a "middle arrow" which is nuclear and does make all the squares commute. So depending on your intended applications, this might be of some use; it says we can manufacture an "extension" of $T_1$ and $T_3$ which is nuclear. Furthermore, even if one wants to show that a given $T_2$ is nuclear, this construction might help since under certain extra conditions one may be able to prove that $\theta=T_2$.