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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 $\sqrt{\left(x_1-(2+x_1)\right)^2+\left(x_2-(-x_2)\right)^2+...+\left(x_{n-1}-(-x_{n-1})\right)^2}$ $=2\sqrt{1+x_2^2+...+x_{n-1}^2}$. 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 in a direction parallel to one of the edges of that facet (and nowhere else): the minimizing path follows that direction all the way.

This answer realizes the idea from the answer by Gerhard "Mentally Less Energetic" Paseman.

NB: as pointed out by the OP in the comment below, this answer is incomplete. I will think about adding other possible unfoldings into consideration.

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 $\sqrt{\left(x_1-(2+x_1)\right)^2+\left(x_2-(-x_2)\right)^2+...+\left(x_{n-1}-(-x_{n-1})\right)^2}$ $=2\sqrt{1+x_2^2+...+x_{n-1}^2}$. 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 in a direction parallel to one of the edges of that facet (and nowhere else): the minimizing path follows that direction all the way.

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 in a direction parallel to one of the edges of that facet (and nowhere else): the minimizing path follows that direction all the way.

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 $\sqrt{\left(x_1-(2+x_1)\right)^2+\left(x_2-(-x_2)\right)^2+...+\left(x_{n-1}-(-x_{n-1})\right)^2}$ $=2\sqrt{1+x_2^2+...+x_{n-1}^2}$. 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 in a direction parallel to one of the edges of that facet (and nowhere else): the minimizing path follows that direction all the way.

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).

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 in a direction parallel to one of the edges of that facet (and nowhere else): the minimizing path follows that direction all the way.