First, let $k^{2n}=a_k p + r_k$ with $0<r_k<p$. It is clear that, $$ \sum_{k=1}^{\frac{p-1}{2}}\left\{\frac{k^{2n}}{p}\right\}= \sum_{k=1}^{\frac{p-1}{2}} \frac{r_k}{p}. $$Next, we try to select a prime for which the computation of the sum of $r_k$'s is easy. For this goal, we investigate when the map, $x\mapsto x^{2n}\pmod{p}$ is injective over $\{1,2,\dots,\frac{p-1}{2}\}$. Take $x,y\in \{1,2,\dots,\frac{p-1}{2}\}$, and let $x^{2n}\equiv y^{2n}\pmod{p}$. Using Fermat's theorem, we also have, $x^{p-1}\equiv y^{p-1}\pmod{p}$. Hence, if we select $p$ such that $p-1$ and $n$ are coprime, we have that $r_k$ sweeps all quadratic residues. Now, it suffices to select $p\equiv 2\pmod{n}$ for this goal. Next, if $a$ is a quadratic residue modulo $p$, and $p\equiv 1\pmod{4}$, then $p-a$ is also a quadratic residue. Hence, if we let $\mathcal{R}$ to be the set of all $\frac{p-1}{2}$ non-zero quadratic residues in modulo $p$, we have $a\in \mathcal{R}\iff p-a\in \mathcal{R}$. Therefore, $\sum_{a\in\mathcal{R}}=\sum_{a\in \mathcal{R}}p-a$, together with $\sum_{a\in \mathcal{R}}a + \sum_{a\in\mathcal{R}}p-a = p\frac{p-1}{2}$ yields that, $\sum_{k=1}^{\frac{p-1}{2}}r_k = p\frac{p-1}{2}$. From here, if $p\equiv 2\pmod{n}$ and $p\equiv 1\pmod{4}$, we immediately conclude that, the given arithmetic ratio is precisely $\frac12$. Finally, by Dirichlet's theorem on arithmetic progressions, we have infinitely many $p$ for which $p\equiv 2\pmod{n}$ and $p\equiv 1\pmod{4}$.

Post for storage: construct bijection a->p-a with p=1[4], now if x^n cover all residue mod p then arithmetic mean be 1/2. It means (p-1, n)=1, which is infinitely exist by Dirichlet theorem, done!

Am I wrong ?? (I didn't use dirichlet) Just let p be any prime which is 1 mod 4. Then let $k^2$ be congruent to r(k) mod p. Then r(k) is a q.r. and by legendre so is -r(k). Now we can easily see that r(k) are distinct and so they form all quadratic residues mod p. Now we can pair this(even) number of element by r(k) with -r(k) and we can easily see that this pairs are disjoint. Now the sum $r(1)^n$+... can be grouped in (p-1)/4 groups,each with sum p(because the residues are not 0 mod p and their sum is divisible by p and less then 2p) so the final answer will be $p*(p-1)/4$ divided by $p*(p-1)/2$ so the answer will be 1/2 .

Does anyone have an idea how to prove the general case (if it's even correct)?

so basically it's enough to find another constant except $\frac{1}{2}$ or to prove the problem without $p\equiv 1\pmod 4$.

Bump.