@DavidLoeffler That's very true, and it turns out that not for many integers $n$, the expression $8n/\pi^2$ is an integer. :-)

I'm confused about the meaning of "modular" in the statement of the question. For example, if you have a polarized Abelian surface, then you expect that there is a Siegel modular form (not necessarily cuspidal) whose L-functions are related to the Hasse-Weil zeta function of the Abelian surface. If the geometric endomorphism ring is Z, then you expect to see a stable cuspidal Siegel modular form; whereas, if the geometric endomorphism ring is non-commutative, then you can prove that the Siegel modular form is going to be an Einsenstein series coming from the Klingen parabolic subgroup.

In particular, if $n$ is even, which $p-1$ is, you always have $0.6 n \leq f(n) \leq 0.75 n$. If $n$ is odd, you get $8n/\pi^2 \leq f(n) \leq 8n/9$.

But suppose $f$, $g$, and $h$ are holomorphic modular forms. Is there a way to prove (potential) automorphy of $f \times g \times h$?

By Cogdell--Piatetski-Shapiro, j automorphic on GL(n) for n up to 6 is sufficient. So, the long answer is no as well.

Thanks everyone. You can actually count the number of subrings of index $p$. Namely, if $r(x) \mod p$ has $u$ factors of degree $1$ and $w$ factors of degree $2$, and the rest of higher degrees, then the number of subrings of ${\mathbb Z}/p{\mathbb Z}[x]/(r(x))$ of index $p$ is ${u \choose 2} + w$.

In our work we only need this for unramified primes.

@KConrad. yes, it does, for example in Evanston, at 11:24 PM, sitting at a bar with Nathan Kaplan.