This answer realizes the idea from the answer by Gerhard "Mentally Less Energetic" Paseman.
The unfolding of the surface of the $n$-cube into the ($n-1$)-space is the union of $2n$ ($n-1$)-cubes (facets of the $n$-cube); they may be arranged in this fashion: one central ($n-1$)-cube, with vertices $(\pm\frac12,\pm\frac12,...,\pm\frac12)$, its shifts by $1$ and by $-1$ along each of the coordinate axes, and one more shift by $2$ in one of these directions.
We actually only need two among these, the central cube and one of its adjacent cubes, since they already contain representatives of the opposite vertices. These are e. g. $(-\frac12,-\frac12,...,-\frac12)$ and $(1,0,...,0)+(\frac12,\frac12,...,\frac12)=(\frac32,\frac12,...,\frac12)$.
The distance thus is $\sqrt{(\frac32-(-\frac12))^2+(n-2)(\frac12-(-\frac12))^2}=\sqrt{n+2}$. This is the shortest distance in question.