It is quite likely. At least, the proof for the case $d\mid n$ is easy. First of all, the restriction $i\ne j$ does not matter: adding $n$ ones changes nothing in the problem. Now notice that $|\langle v_i,v_j\rangle|\ge \langle v_i,v_j\rangle^2=\langle V_i,V_j\rangle$ where $V_i=v_i\otimes v_i$. Now, $\langle V_i,I\rangle=1$ for all $i$ ($I$ is the identity matrix, as usual), so $\langle\sum_i V_i,I\rangle=n$ and, by Cauchy-Schwarz, $\|\sum_i V_i\|^2\ge n^2/\|I\|^2=n^2/d$ (the norm here is the Frobenius norm, i.e., the square root of the sum of the squares of the matrix elements), which results in $\min_{v_i}\sum_{i,j}\langle v_i,v_j\rangle^2\ge n^2/d$. For the conjectured minimizer, both this estimate and the crude inequalities $|\langle v_i,v_j\rangle|\ge \langle v_i,v_j\rangle^2$ become identities, whence the conclusion.

I do not see off hand how to modify this argument for the case $d\not\mid n$ but it still makes the conjecture quite plausible. In the worst case scenario, you are off by at most $d/4$ from the true minimum with your system.

1more comment