If $a = 0$ then the first equation implies $b^{n+2}c = 0$. Hence $b = 0$ or $c = 0$; w.l.o.g., $b = 0$. Then the last equation reduces to $c^{n+3} = 0$ which implies $c = 0$. Thus if $a = 0$ then $a = b = c = 0$. Analogously we can prove that if $b = 0$ or $c = 0$ then $a = b = c = 0$. The triple $(0,0,0)$ satisfies the equation.

It remains to consider triples $(a,b,c)$ whose all terms are different from $0$. Let $p$ be an arbitrary prime number. If $p \mid a$ then the first equation implies $p \mid b^{n+2}c$. Thus $p \mid b$ or $p \mid c$; w.l.o.g., $p \mid b$. Then the last equation gives $p \mid c^{n+1}$ which implies $p \mid c$. Thus if $p \mid a$ then $p \mid a, p \mid b, p \mid c$. Analogously we can prove that if $p \mid b$ or $p \mid c$ then $p \mid a, p \mid b, p \mid c$.

But rewriting $a = pa', b = pb', c = pc'$ enables us to divide both sides of all equations by $p^{n+3}$, giving a similar system of equations having $a', b', c'$ in the role of $a, b, c$. Hence the triple $(a', b', c')$ also satisfies the system of equations. As the numbers other than $0$ cannot be infinitely divided in integers, a finite number of divisions should give us a solution whose terms have no common prime factors. By the previous paragraph, this is possible only if the values of variables have no prime factors, i.e., $a = \pm 1, b = \pm 1, c = \pm 1$. We show now that such solutions do not exist. Indeed, if $n$ is even then the system of equations reduces to
$$\begin{cases} a + c + c + b = 0, \\ b + a + a + c = 0, \\ c + b + b + a = 0. \end{cases}$$
Adding all equations results in $4(a + b + c) = 0$, implying $a + b + c = 0$; but the sum of three odd numbers cannot equal the even number $0$.

If $n$ is odd then the system of equations reduces to
$$\begin{cases} 1 + bc + 1 + ab = 0, \\ 1 + ca + 1 + bc = 0, \\ 1 + ab + 1 + ca = 0. \end{cases}$$
Adding all equations results in $2(ab + bc + ca) + 6 = 0$, implying $ab + bc + ca = -3$. As $|ab| = |bc| = |ca| = 1$, the only possibility is $ab = bc = ca = -1$. This demands that $a, b, c$ have pairwise opposite signs which is impossible. Consequently, no solutions except $(0,0,0)$ exist.