If I recall correctly, the duality functor $D: {\sf Ban}_1 \to {\sf Ban}_1^{\rm op}$ is a left adjoint -- the analogous statement for ${\sf Vect}$ and ${\sf Vect}^{\rm op}$ is in Mac Lane's CFTWM somewhere -- and so $D$ should preserve colimits. That is, if we think of it as a contravariant functor on ${\sf Ban_1}$, it should take colimits to limits.

However, you are asking about duals of limits (in the category-theoretic sense) and so one would not expect the dual of a limit to be the colimit of the duals. For instance, the product in ${\sf Ban}_1$ of countably many copies of the ground field is $\ell_\infty$, whose dual is (assuming Axiom of Choice in some form) not the same as $\ell_1$ (and $\ell_1$ is the coproduct of countably many copies of the ground field.

Actually, I think for your particular question we have a counterexample which doesn't use AC/Hahn-Banach. Consider an "inverse system" $\dots E_n \to E_{n-1} \to \dots E_0$ where $E_n=\ell_1^n$ and the linking maps are given by "omit the last entry of the vector". Then the inverse/projective limit should be $\ell_1$, whose dual is $\ell_\infty$. But the dual of the inverse system would be $E_0^* \to E_1^*\to\dots E_n^* \to \dots$ and here $E_n^*=\ell_n^\infty$ with linking maps given by obvious inclusion. The direct/inductive limit of this system will be $c_0$, not $\ell_\infty$.