Let $G$ be a connected split reductive group over $\mathbb{Z}$. Let $n$ be a positive integer. Let $i_n(q)$ be the number of elements of $G(\mathbb{F}_q)$ satisfying $x^n=1$.
Question: Is there a "nice" case-free formula for $i_n(q)$? Is it always a polynomial in $q$?
The "best hope" is a formula akin to Steinberg's formula for the number of elements of $G(\mathbb{F}_q)$ (cf. Section 7.3 of these notes). Can this best hope be realised?
For instance, suppose $n=2$. Then we are considering the number of involutions in finite reductive groups. A case by case analysis shows that there exists polynomials $P, R\in \mathbb{Z}[t]$ such that $i_2(q)=P(q)$ for all even $q$ and $i_2(q)=R(q)$ for all odd $q$, cf. this paper. However, I do not know a case-free proof of this fact. Nor do I know a case free expression for these polynomials.
Remark 1. If it helps, we can assume that the characteristic is large enough.
Remark 2. Aforementioned result of Steinberg amounts to giving an expression for $|\mathrm{Hom}(\mathbb{Z}, G(\mathbb{F}_q)|$. The above question asks for an expression for $|\mathrm{Hom}(\mathbb{Z}/n, G(\mathbb{F}_q)|$.