Suppose $n$ is $3m$. Take the points to be
$$(1, 0, 0), (2, 0, 0), \dots, (m, 0, 0), \\ (0, 1, 0), (0, 2, 0), \dots, (0, m, 0), \\ (0, 0, 1), (0, 0, 2), \dots, (0, 0, m)$$
Take $P$ to be $(x, y, z)$. Then the sum is $2m|x| + |x-1| + |x-2| + \dots + |x-m|$ plus similar terms in $y$ and $z$. Evidently we can minimize for $x, y, z$ separately. Looking at the terms in $x$, it is clear that if we increase $|x|$ by $k > 0$, then we increase $2m|x|$ by $2mk$. At most we can reduce the other terms by $mk$, so we increase the sum by at least $mk$. Similarly, if we reduce $|x|$ by $k > 0$, we reduce the sum by at least $mk$. So to minimize the sum we must take $|x| = 0$. Similarly for $y$ and $z$. Hence $P$ is at the origin which is outside the convex hull.
If $n$ is $3m-1$ or $3m-2$, then we drop $(0, 0, m)$ and/or $(0, m, 0)$. Then the argument is strengthened for the $y$ and $z$ coordinates. For the $x$ coordinate we have at worst $(2m-2)|x| + |x| + |x-1| + \dots + |x-m|$. If $n > 6$, then $m > 2$, then $2m-2 > m$ and the same argument works. If $n = 5$, then we have $3|x| + |x| + |x-1|$ for the $x$-coordinate, and the same argument works.