Finding all vertices of the polytope would have the same complexity as the Vertex Enumeration Problem. I do not think that's a practical approach.

The polytope is the intersection of the box $0\leq x \leq v$, and the hyperplanes $Ax=b$. A point in the interior of the box, which also lies on the hyperplanes is what is desired. Consider the following convex optimization problem:
$$
\max_{x\in R^n} \left\{\min \{x_1, v_1-x_1,\cdots, x_n, v_n-x_n\} \right\} ~\mbox{subject to}~ Ax=b ~~\&~~ 0\leq x\leq v.
$$

The rationale is that the numbers within the $\min$ brackets are the distances from the $2n$ hyperplanes defining the box. Thus, maximizing the minimum of those forces the optimal point to be in the interior. The optimization problem can be written as an LP with $n+1$ variables and $m+4n$ constraints. Hence this gives a polynomial time algorithm for determining a point in the interior. Also note that the optimal value is positive if and only if such an interior point exists.