Answer. The smallest possible integer with that property is $n = 13$.

We note that we have $xyz \mid (x+y+z)^n$ if and only if for each prime $p$ the inequality $v_p(xyz) \le v_p((x+y+z)^n)$ holds, where as usual $v_p(m)$ denotes the exponent of $p$ in the prime factorization of $m$.
Let $x$, $y$ and $z$ be positive integers with $x \mid y^3$, $y \mid z^3$ and $z \mid x^3$. Let $p$ be an arbitrary prime, and w.l.o.g. let the multiplicity of $p$ be lowest in $z$, that is, $v_p(z) = \min\{v_p(x), v_p(y), v_p(z)\}$.
Then we have $v_p(x + y + z) \ge v_p(z)$, and from the divisibility constraints we get $v_p(x) \le 3v_p(y) \le 9v_p(z)$. It follows that
$$\begin{aligned} v_p(xyz) &= v_p(x) + v_p(y) + v_p(z) \\ &\le 9v_p(z) + 3v_p(z) + v_p(z) = 13v_p(z) \\ &\le 13v_p(x + y + z) = v_p((x + y + z)^{13}), \end{aligned}$$
which proves that for $n = 13$ the desired property is satisfied.
It remains to show that this is indeed the smallest possible integer with this property. For doing so, let $n$ now be a number that has the desired property. By setting $(x, y, z) = (p^9, p^3, p^1)$ with an arbitrary prime $p$ (in order to achieve that both inequalities in the previous calculation become equalities), we get
$$\begin{aligned} 13 &= v_p(p^{13}) = v_p(p^9 \cdot p^3 \cdot p^1) = v_p(xyz) \\ &\le v_p((x + y + z)^n) = v_p((p^9 + p^3 + p^1)^n) = n \cdot v_p(p(p^8 + p^2 + 1)) = n, \end{aligned}$$
which yields $n \ge 13$.