The elementary symmetric polynomials appear when we expand a linear factorization of a monic polynomial: we have the identity $$ \prod _{j=1}^{n}(\lambda -X_{j})=\lambda ^{n}-e_{1}(X_{1},\ldots ,X_{n})\lambda ^{n-1}+e_{2}(X_{1},\ldots ,X_{n})\lambda ^{n-2}+\cdots +(-1)^{n}e_{n}(X_{1},\ldots ,X_{n}). $$
Let $\vec v\in \mathbb Z^n$, where the first element is always $0$, i.e. $e_1(...)=0$. How many solutions for $X_k\in\mathbb C$ do we get for the following: $$\vec v=\pmatrix{e_1\\e_2\\\vdots\\e_n}?$$
Some low dimensional numerical experiments point toward $n!$. Is this true?
No, this is not true. The condition on $e_1$ is irrelevant, as is the condition that the entries of $v$ are integers. You're asking how many ways there are to order the roots of some polynomial of degree $n$, and the answer depends on their multiplicities. If the roots each have multiplicity $1$ then the answer is $n!$ but, for example, if there's only one root with multiplicity $n$ then the answer is $1$. In general if the roots have multiplicities $m_i$ then the answer is
$$\frac{n!}{\prod m_i!}.$$