I claim that this is false for $n=6$. I find it convenient to shift the variables to $w_{ij} = 2 v_{ij} -1$. So the inequalities are $$-1 \leq w_{ij} \leq 1 \quad (1)$$ $$w_{ij} + w_{ji}=0 \quad (2)$$ $$-1 \leq w_{ij} + w_{jk} + w_{k i} \leq 1. \quad (3)$$
Consider the point $x_{12} = x_{34} = x_{56} = 1$, $x_{23} = x_{45} = x_{61} = -1$ and $x_{ij}=0$ in all cases that are not forced from the above by skew symmetry. I claim that $x$ is a vertex.
Inequalities $(1)$ and $(2)$ are clear. For (3), if $(i,j,k)$ are cyclically consecutive modulo $6$, then the sum in the middle is $0$. Otherwise, at most one of the summands is nonzero, so the inequality is clear.
We now must verify that $x_{ij}$ is a vertex. If not, it is in the interior of a line segment, so there is some vector $e_{ij}$ so that $x_{ij} \pm e_{ij}$ is inside the polytope for both choices of sign. Since we must preserve the truth of $(2)$, we have $e_{ij} + e_{ji} = 0$. In order to have $x_{12} \pm e_{12}\leq 1$ for both choices of sign, we must have $e_{12}=0$. Similarly, $e_{i(i+1)}=0$ (with indices cyclic modulo $6$.)
Now, we have $x_{12} + x_{24} + x_{41} = 1$. In order to have $(x_{12} \pm e_{12}) + (x_{24} \pm e_{24}) + (x_{41} \pm e_{41}) \leq 1$ for both choices of sign, we must have $e_{12} + e_{24} + e_{41} =0$. And we already know $e_{12}=0$. So $e_{24} + e_{41}=0$. We can get $12$ linear equations in this manner: $e_{i(i+2)} + e_{(i+2)(i-1)} = 0$ and $e_{i(i+3)} + e_{(i+3) (i+5)}=0$ (indices cyclic modulo $6$). Combined with $e_{ij}=-e_{ji}$, the only solution is that all of the $e$'s are zero.
So $x$ is not at the midpoint of any line segment in the polytope, and is a vertex.
Here is a more conceptual explanation. Let $x$ be a point in the polytope. Let $\Gamma(x)$ be the graph with vertex set $[n]$ and an edge $(i,j)$ if $|x_{ij}|=1$. Let $\Delta$ be the $2$-dimensional simplicial complex with a face $(i,j,k)$ if $|x_{ij}+x_{jk} + x_{ki}|=1$. Let $e_{ij}$ be a potential perturbation of $x_{ij}$, as in the solution. If $H^1(\Delta, \mathbb{R})=0$, then inequalities $(3)$ force there to be constants $a_i$ so that $e_{ij} = a_i - a_j$ for every edge $(i,j) \subset \Delta$. If $\Gamma \subset \Delta$, then we must also have $a_i = a_j$ whenever $i$ and $j$ are in the same connected component of $\Delta$. So, if we can arrange that $\Gamma$ is connected, $H^1(\Delta)=0$ and $\Gamma \subset \Delta$, then we are at a vertex. This was the sort of heuristic I used to find the above example.