When does this system of congruences hold?

Galois theory of finite fields sheds some light on this question. In fact, it follows that the latter congruence will always be a consequence of the former! Basically because they are each others Galois conjugates!

As you observed $\alpha^{\pm1}=(n\pm\sqrt{n^2-4})/2$ are the solutions of the quadratic equation $$ m(x)=x^2-nx+1=0. $$ Everything takes place in the ring $R$ of algebraic integers of $\Bbb{Q}(\sqrt{n^2-4})$. From basic algebraic number theory we infer that $R$ has a prime ideal $\mathfrak{p}$ such that $\mathfrak{p}\cap\Bbb{Z}=p\Bbb{Z}$. The quotient ring $R/\mathfrak{p}$ is then a finite field $K$. If $n^2-4$ is a quadratic residue modulo $p$ then $|K|=p$. But if $m(x)$ is irreducible modulo $p$ then $|K|=p^2$.

The idea is that for integers a congruence modulo $p$ is equivalent to congruence modulo $\mathfrak{p}$, and the latter can be decided by projecting everything to the quotient field $K$.

If $m(x)$ factors modulo $p$, then the images of $\alpha^{\pm1}$ in $K$ are residue classes of integers modulo $p$. So Little Fermat says that $\alpha^p\equiv\alpha\pmod{\mathfrak{p}}$ and also $\alpha^{-p}\equiv\alpha^{-1}\pmod{\mathfrak{p}}.$ Consequently $$\alpha^p+\alpha^{-p}\equiv\alpha+\alpha^{-1}\pmod{\mathfrak{p}}$$ and this implies the same congruence of integers modulo $p$.

If $m(x)$ is irreducible modulo $p$ then its zeros in $K$ are Frobenius conjugates of each other. As those zeros are the projections of $\alpha^{\pm1}$, it follows that $$ \alpha^p\equiv\alpha^{-1}\pmod{\mathfrak{p}}. $$ Applying the Frobenius again then gives the congruence $$ \alpha^p+\alpha^{-p}\equiv\alpha^{-1}+\alpha\pmod{\mathfrak{p}} $$ and we are done by repeating the earlier argument.

The conclusion is that the condition $$n_1+n_2\equiv n_3\pmod p$$ is all you need for both of the congruences to hold.