How to show that $[L_i, v_j]=i\hbar\sum_k \epsilon_{ijk}v_k$ for any vector $\textbf{v}$ constructed from $\textbf{x}$ and/or $\nabla$?

It follows from the fact that $\hat L_i$ are generators of the rotations.

Generally, the vector operator $\hat V_i$ is not just "a list" of operators. It has to transform like a vector under rotations $\hat V_i' = R_{ij} \hat V_j$ (summation implied).

This identity can be proven by considering an observable in a rotated frame of reference. The frame rotation can be carried out through a transformation $R$ acting on $\hat V$ or a unitary operator $\hat{U}(R)$: \begin{equation} \left._R\right. \langle\alpha|\hat{V}_i|\beta\rangle_R = \langle\alpha|\hat U^\dagger \hat V_i \hat U|\beta\rangle = R_{ij}\langle\alpha| \hat V_j |\beta\rangle \end{equation}

Thus, we have: $\hat U^\dagger \hat V_i \hat U = R_{ij} \hat V_j$. By considering infinitesimal rotation $\hat U = 1 - \frac{i \epsilon \bf{J}\cdot \bf{n}}{\hbar}$ and corresponding $R = \left(\begin{matrix}1& -\epsilon & 0\\ \epsilon & 1 &0 \\ 0 & 0 & 1\end{matrix}\right)$ we find at first order in $\epsilon$: \begin{equation} [V_i, J_j] = i \hbar \epsilon_{ijk} V_k \end{equation}

A nice proof of this identity can be found here:

Similar question: Angular and linear momentum operators' commutation

Another useful reference: