For $a=2$, $b=c=\frac{1}{2}$ and $n \ge 3$ the inequality is not satisfied. On the other hand, for $n=1$ it is equivalent to the AM-GM inequality. It remains to consider the case $n=2$.
We shall prove that the inequality is true for $n=2$.

First solution. Set $x = bc$. Then
$$abc(a^2 + b^2 + c^2) = ax(a^2 + (b+c)^2 - 2x).$$
The function $x(p-2x)$ is increasing for $x \le \frac{p}{4}$. Since
$$bc \le \frac{(b+c)^2}{4} \le \frac{a^2 + (b+c)^2}{4},$$
it follows that if $b+c = b'+c'$ and $bc \le b'c'$, then
$$abc(a^2 + b^2 + c^2) \le ab'c'(a^2 + b'^2 + c'^2).$$
Without loss of generality we can assume that $b \le 1 \le c$. Set $b' = 1, c' = b+c-1$. Since $b+c = b'+c'$ and $bc-b'c' = (b-1)(c-1) \le 0$, the above implies that
$$abc(a^2 + b^2 + c^2) \le a(2-a)(a^2 + 1 + (2-a)^2).$$
Set $d = (a-1)^2$. Then
$$a(2-a)(a^2+1+(2-a)^2) = (1-d)(3+2d) = 3-d-2d^2 \le 3$$
and the given inequality (for $n=2$) follows.

Second solution. Let $n=2$ and set $ab+bc+ac=x$. Then we have $a^2+b^2+c^2=9-2x$ and $9abc \le x^2$ (this follows from the inequality $(ab+bc+ac)^2 \ge 3abc(a+b+c)$). Hence we have to prove that $(9-2x)x^2 \le 27$, which is equivalent to the obvious $(2x+3)(x-3)^2 \ge 0$.