I claim the answer is $1$ and powers of three. For $n=1$ $p$ is clearly irreducible with roots being non-real cube roots of $1$ and the case $n=3^k$ follows by Eisenstein on $p(x+1)$. For any $n \geqslant 2$ not divisible by $3$ and $z$ a non-real cube root of $1$ the polynom $p(z)$ clearly has roots are the cube roots, hence it is divisible by $1+x+x^2$ and cannot be irreducible. Finally if $n = b 3^a $ for $3 \nmid b$, then $p(x) = \frac{x^{b 3^{a+1}} - 1}{x^b - 1}$ and is zero for any non-real $3^{a+1}$-th root of $1$, hence divisible by $x^{2a} + x^a + 1$.

Hmm, what about $n=2$: $$x^4+x^2+1 = (x^2+1)^2-x^2$$