.5${}{}{}{}{}{}{}$
Let me atone for giving too few details by giving too many.
Let $$S=\sum_{a_1=0}^{p-1}\dots\sum_{a_n=0}^{p-1}\sum_{t=0}^{p-1}\sum_{m=0}^{p-1}e^{2\pi im(a_1^2+\cdots+a_n^2-t^2)/p} $$ The innermost sum is $p$ if $a_1^2+\cdots+a_n^2-t^2\equiv0\pmod p$ and zero otherwise, so $S$ counts $2p$ whenever $a_1^2+\cdots+a_n^2$ is a (nonzero) quadratic residue, $p$ whenever it's zero. On the other hand, $$ S=\sum_{m=0}^{p-1}\sum_{a_1=0}^{p-1}\dots\sum_{a_n=0}^{p-1}\sum_{t=0}^{p-1}e^{2\pi im(a_1^2+\cdots+a_n^2-t^2)/p} $$ so $$ S=p^{n+1}+\sum_{m=1}^{p-1}\sum_{a_1=0}^{p-1}\dots\sum_{a_n=0}^{p-1}\sum_{t=0}^{p-1}e^{2\pi im(a_1^2+\cdots+a_n^2-t^2)/p} $$ so $$ S=p^{n+1}+\sum_{m=1}^{p-1}\left(\left(\sum_{a_1=0}^{p-1}e^{2\pi ima_1^2/p}\right)\cdots\left(\sum_{a_n=0}^{p-1}e^{2\pi ima_n^2/p}\right)\left(\sum_{t=0}^{p-1}e^{2\pi imt^2/p}\right)\right) $$ Each of those inner sums is a Gauss sum and known to equal $\sqrt p$ in modulus (more detail: the sum is ${m\overwithdelims()p}\sqrt{{-1\overwithdelims()p}p}$), so $|S-p^{n+1}|\le(p-1)p^{(n+1)/2}$. For $n\gt1$, the main term beats the error term, and you get a good estimate.