Somewhat relevant: arxiv.org/abs/0811.0876

The unique (up to conjugation) normalized eigenform in $S_2(\Gamma_0(23))$ has coefficients in $\mathbf{Q}(\sqrt{5})$, and the corresponding mod-$2$ representation lands in $S_3$ (the representation comes from the Hilbert class field of $\mathbf{Q}(\sqrt{-23})$). It follows that all the $a_p$ with $p$ prime lie in $\mathbf{Z}[\sqrt{5}]$ and their mod $2$ reduction lands in $\mathbf{F}_2$. But $a_2 = (-1 + \sqrt{5})/2$, so $a_2 \mod 2$ generates $\mathbf{F}_4$.

@Ventullo: Yes it does (oops). In my defense, the KW-argument I give doesn't require Ihara's Lemma (the proof goes through Taylor's Ihara avoidance), and so generalizes well to higher rank groups.

Dear Noam, exactly correct, although since I'm lazy, it was much easier just to look through the Cremona tables. For comparison, the first four primes which split completely in $\mathbf{Q}(E[3])$ are $61$, $313$, $349$, and $373$. (For some reason, I missed the curve of conductor $1952$ on my first try.)