It is more or less easy to obtain, via exponential sums, an asymptotic formula for the number $J$ of solutions of the congruence

$$x^{n}+y^{n} \equiv z^{n} \pmod{p}$$

where $$1 \leq x, y, z \leq p-1.$$

Indeed, since

\begin{eqnarray*} J &=& \sum_{x=1}^{p-1} \sum_{y=1}^{p-1} \sum_{z=1}^{p-1} \frac{1}{p}\sum_{a=0}^{p-1}e^{2 \pi i \frac{a(x^{n}+y^{n}-z^{n})}{p}}\\
&=& \frac{(p-1)^{3}}{p}+ \frac{1}{p} \sum_{a=1}^{p-1} \sum_{x=1}^{p-1}\sum_{y=1}^{p-1} \sum_{z=1}^{p-1}e^{2\pi i \frac{a(x^{n}+y^{n}-z^{n})}{p}}\\
&=& \frac{(p-1)^{3}}{p} + \frac{1}{p}\sum_{a=1}^{p-1}\left|\sum_{x=1}^{p-1}e^{2\pi i \frac{ax^{n}}{p}}\right|^{2}\sum_{y=1}^{p-1}e^{2\pi i \frac{ay^{n}}{p}}\\
\end{eqnarray*}

and

$$ \sum_{a=0}^{p-1} \left| \sum_{x=1}^{p-1} e^{2\pi i \frac{ax^{n}}{p}} \right|^{2} \leq n\,p\,(p-1)$$

and

$$\left|\sum_{y=1}^{p-1}e^{2\pi i \frac{\alpha y^{n}}{p}}\right| \leq n\sqrt{p}$$

for any integer $\alpha$ which is not divisible by $p$, it follows that

$$J = \frac{(p-1)^{3}}{p}+O\left(n^{2}(p-1)\sqrt{p}\right).$$