This is not an answer, just a remark.
I find your question interesting even for small values of $n=|V|$.
For example, if $n=4$, then choosing $V$ as the vertices of a regular tetrahedron,
the $\binom{4}{2}=6$ planes $V_P$ determine six normal lines that define a cuboctahedron:
The angles between these lines/planes is either $90^\circ$ or
about $63.6^\circ$ (or $116.4^\circ$) $60^\circ$ (or $120^\circ$).
Now, the optimal packing of six lines is known (Conway, Hardin, Sloane) to be the six diameters of the icosahedron.
The minimum angle determined by those diameters is a bit larger, if I've calculated
correctly: $63.4^\circ$.
So: Is there an arrangement of four vectors $V$ that yields this optimal line packing? Answer: No, $60^\circ$ is the optimal for $n=4$. See Henry Cohn's argument in the comments.
Here is Edmund's suggestion for $n=6$: $V$ is given by half the vertices of an icosahedron (blue), which generate 15 normal lines (red) passing through the midpoints of the icosahedron's 30 edges, with the normal lines separated by $49.7^\circ$.
The tips of the normal lines form the vertices of an icosidodecahedron.
