So your question is as follows: let $n$ be a positive integer, $q$ the smallest prime such that $q\sharp > n$, $e>q$ be an integer and $r$ be an $e$-th power mod $n$. Then is $r$ a $p$-th power for some $p\leq q$?
In general, no. Take a large prime $q$ and $n$ be the greatest power of $2$ less than $q\sharp$: then $n \geq e^{(1+o(1))q}$ by the PNT, so we can find (PNT again) some prime $s>q$ such that $2^s < n$. Take $r=2^s$ and $e=s$, assume that $r$ is a $p$-th power for some $p \leq q$. But this implies that the $2$-adic valuation of $r$ is divisible by $p$, which is impossible.
Now, let’s state a simple condition under which the statement holds (albeit for mostly trivial reasons).
Note that $q\sharp= e^{\theta(q)} \in [c^{-1},c]e^q$ where $c=e^{0.007q/\ln{q}}$ (by eg this): so if $n \geq 6$, we see that $n \leq q\sharp < 3^q$.
Our assumption is that we also have $v_2(n) \leq q$.
If $e$ has a prime divisor $p \leq q$, then $p$ works. Otherwise, replacing $e$ with one of its prime divisors, we can assume that $e$ is prime.
As $\varphi(n) < n \leq q\sharp$, there is some $p\leq q$ not dividing $\varphi(n)$, so that taking the $p$-th power in $(\mathbb{Z}/d\mathbb{Z})^{\times}$ is an isomorphism for all $d |n$.
Moreover, if $s$ is a prime power dividing $n$ and coprime to $n/s$, such that $r$ and $s$ are not coprime, then the valuation of $s$ is at most $q$ (by the assumption), so the only noninvertible $e$-th power mod $s$ is zero (every noninvertible element to the $e$-th power is zero). Thus $r$ is zero mod $s$ (hence a $p$-th power).
It follows by CRT that $s$ is a $p$-th power mod $n$, QED.
