Yes. In fact, if you work in $\mathsf{ZF}+\mathsf{DC}+$ "all sets of reals have the property of Baire" ($\mathsf{BP}$), say the Solovay or Shelah models, you can prove that $(\ell^\infty)^* = \ell^1$, so that the only continuous linear functionals on $\ell^\infty$ are those coming from $\ell^1$. The generalized limits are certainly not of this form, since they vanish on $c_0$ but not identically. Under these axioms, $\ell^1$ is reflexive.

You can find a proof in Schechter, *Handbook of Analysis and its Foundations*, sections 29.37–38.

Interestingly, under these axioms all linear functionals on any Banach space are continuous, so in fact the *algebraic* dual of $\ell^\infty$ also equals $\ell^1$. And the limit operator has no linear extension to $\ell^\infty$.

Digression: This is a fun example because there are several different ways to prove in $\mathsf{ZFC}$ that $(\ell^\infty)^* \ne \ell^1$. It is interesting to see how all of them fail in this model, and where they used choice. For instance:

Use Hahn-Banach as in your example. Hahn-Banach doesn't work without choice.

Use Alaoglu's theorem to produce a weak-* limit point of the usual basis $\{e_i\}$ of $\ell^1 \subset (\ell^\infty)^*$. This is another sort of generalized limit. So Alaoglu also must fail without choice; of course, the usual proof uses Tychonoff.

Recall the following exercise: "If $X^*$ is separable then so is $X$." Note that $\ell^1$ is separable and $\ell^\infty$ is not. But the "exercise" used Hahn-Banach, and isn't necessarily true without choice.