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.
Two general opposite points on the surface can be now represented as a point $(x_1,x_2,...,x_{n-1})$ inside the central cube (i. e. with $-\frac12\le x_i\le\frac12$, $i=1,...,n-1$) and the reflection of $(-x_1,...,-x_{n-1})$ in the hyperplane passing through $(1,0,...,0)$ perpendicularly to the first axis, i. e. the point $(2+x_1,-x_2,...,-x_{n-1})$ inside the cube shifted by $2$ in the positive direction along the first axis.
The distance thus is $\sqrt{(x_1-(2+x_1))^2+(x_2-(-x_2))^2+...+(x_{n-1}-(-x_{n-1}))^2}=2\sqrt{1+x_2^2+...+x_{n-1}^2}$. Choosing different axes for shifts gives the numbers $2\sqrt{1+x_1^2+...+x_{i-1}^2+x_{i+1}^2+...+x_{n-1}^2}$, and shortest possible distance is the smallest among these numbers.
So evidently $2$ is the absolute minimum, attainable on arbitrary shifts of the center of a facet along one of the axes parallel to that facet (and nowhere else).