First of all, $a=b=c=0$ and $a=b=c=2^{-1/(n-1)}$ are always solutions and, for odd $n$, $a=b=c=-2^{-1/(n-1)}$ is a solution. If any two variables are positive, then they all are. If $n$ is even, then other solutions must be purely positive. If $n$ is odd, the solutions other than $(0,0,0)$ come in opposite pairs. Consider a solution with $0<a\leq b\leq c$. Note that the equations of the system imply that $a=b^n+c^n\geq a^n+b^n=c$. This implies $a=b=c$. So there are no more solutions besides the obvious ones mentioned.