Illustrating @MTyson's construction for $n=2$. We start with an edge $(0,1)$ for the $1$-cube, and replace $(0,1)$ with $(0,0),(1,1)$ and $(1,1),(1,0)$. This leaves one final long diagonal $(1,0),(0,1)$:

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          [![Cuben2][1]][1]

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And the length of the path is $$ 2^1 \sqrt{2} + (2^1 -1 ) \sqrt{1} = 2 \sqrt{2} + 1 \;. $$

Illustrating @MTyson's construction for $d=n=2$. We start with an edge $(0,1)$ for the $1$-cube, and replace $(0,1)$ with $(0,0),(1,1)$ and $(1,1),(1,0)$. This leaves one final long diagonal $(1,0),(0,1)$:

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          [![Cuben2][1]][1]

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And the length of the path is $$ 2^1 \sqrt{2} + (2^1 -1 ) \sqrt{1} = 2 \sqrt{2} + 1 \;. $$

For $d=n=3$, MTyson's construction yields exactly the path I illustrated: $(1, 7, 2, 8, 3, 5, 4, 6)$. Here I will use the vertex indices illustrated in the main post, rather than the coordinates.

• One starts with the $2$-cube path $(1,2,3)$.
• The $(1,2)$ edge is replaced by $(1,7,2)$.
• The $(2,3)$ edge is replaced by $(2,8,3)$.
• The $(3,4)$ edge is replaced by $(3,5,4)$.
• Finally, the last diagonal is added: $(4,6)$.