It's already false for $n=4$. I claim that $$x := \begin{pmatrix} & 1 & 1/2 & 0 \\ 0 & & 1 & 1/2 \\ 1/2 & 0 & & 1 \\ 1 & 1/2 & 0 & \\ \end{pmatrix}$$ is a vertex.
If $x$ is not a vertex, then it lies in the interior of a line segment. Let the line segment have direction $$\pm \begin{pmatrix} & a & b & c \\ -a & & d & e \\ -b & -d & & f \\ -c & -e & -f & \\ \end{pmatrix}$$ We have written the above matrix to be skew symmetric, because traveling along the line segment must preserve the equalities $v_{ij}+v_{ji}=1$.
Travelling along the line segment in both directions must preserve the inequalities $v_{12}$, $v_{23}$, $v_{34}$, $v_{41} \leq 1$. This gives $a=c=d=f=0$. Then preserving the inequalities $v_{32}+v_{21} \geq v_{31}$, $v_{43}+v_{32} \geq v_{42}$, $v_{14} + v_{43} \geq v_{13}$ and $v_{21}+v_{14} \geq v_{24}$ give $-b \leq 0$, $-e \leq 0$, $b \leq 0$ and $e \leq 0$, so $b=e=0$.