I claim the answer is $n=13$. To achieve, take $(x,y,z)=(p,p^9,p^3)$, where $p$ is a prime. Then $v_p(xyz) = 13$ whereas $v_p(x+y+z)=1$, so $n\ge 13$. Conversely, I now show $n=13$ suffices. To see this, note that if $p\mid x$ is a prime, then $p\mid y,z$, and vice versa. So, it suffices to prove the statement for prime powers. Take $p\mid xyz$ and $m=v_p(x)$, $n=v_p(y)$, and $k=v_p(z)$. We have $m\le 3n$, $n\le 3k$ and $k\le 3m$. Assume $m$ is the smallest. So, $n\le 3k\le 9m$ and $k\le 3m$ yields $v_p(xyz) = m+n+k\le 13m$. At the same time, $v_p(x+y+z)\ge \min\{m,n,k\}=m$, so $n=13$ indeed works.