Shortest paths in the Hypercube We'll ask for lengths of shortest paths in the hypercube which obey dimensional constraints.
Given $n\in\mathbb N$, the hypercube or n-cube is the $n$-fold product of the unit interval (so it's not the Hilbert cube!). It is thus the set
$$F_{n,n}\:=\:\left\{\sum\nolimits^n_{i=1}\beta_ie_i\:\Big\vert\:
\beta_i\in \big[0\,,1\big]\right\}$$
where $\,(e_i)_{1\le i\le n}\,$ denotes the standard ONB of $\mathbb R^n$, equipped with the Euclidean distance. For $1\leqslant d\leqslant n\,$ the union of $\,d$-dimensional faces of the $n$-cube is defined as
$$F_{n,d}:=\left\{\sum\nolimits^n_{i=1}\beta_ie_i\:\Big\vert
\text{ where $d$ coefficients }\beta_i\in \big[0\,,1\big],\text{ the remaining ones live in }\{0,1\}\right\},\\[2.5ex]
\text{then } F_{n,1}\:\subsetneq\:F_{n,2}\:\subsetneq\:\ldots\:\subsetneq\: \,F_{n,n}\,.$$ 

What is the length $\,s(n,d)\,$ of a shortest path between the corner at the origin and the opposite corner at $\,(1,1,\ldots,1)\,$ when the path is constrained to lie in $F_{n,d}\,$?

For $\,F_{n,1}$, thus edges only, and $\,F_{n,n}$, allowing for the straight line, the question is quickly answered by $\,s(n,1)=n\,$ and $\,s(n,n)=\sqrt n\,$:
\begin{matrix}{\begin{matrix}&d\\n&\end{matrix}}&1&2&3&4&\ldots\\
1&1\\
2&2&\sqrt 2\\
3&3&\sqrt 5&\sqrt 3\\
4&4&?&??&2\\
\vdots&&&&&\ddots\end{matrix}
The less obvious entry $\,s(3,2)=\sqrt 5\,$ is furnished by 
this eye-catching answer
to the post "What is the geodesic between two opposite corners of a cube on its surface? [closed]". But what about all the rest?
How can the decreasing (certainly!) sequence
$$\,s(n,1),\;s(n,2),\;\ldots,\;s(n,n-1),\;s(n,n)$$
be determined? I didn't find references, neither @SE  nor in the residual internet. Furthermore I'm not sure about the appropriate tagging of the post, but omitting graph-theory was intended.
 A: The following is a generalization of the "visual proof" referred to in the question.
If $p_i\geq0$ $(1\leq i\leq d)$ and $\sum_{i=1}^d p_i=n$ then $$\ell({\bf p}):=\sqrt{\sum_{i=1}^d p_i^2}$$ is minimal when all $p_i$ are equal. When the $p_i$ are forced to be integers then we should make them as equal as possible. To this end assume
$$n= q d+r, \qquad q\geq1,\quad 0\leq r<d$$
and put
$$p_i:=\left\{\eqalign{&q+1\qquad(1\leq i\leq r)\cr &q\quad\qquad(r+1\leq i\leq d)\ .\cr}\right.$$
I claim that
$$s_{n,d}=\ell({\bf p})=\sqrt{dq^2+r(2q+1)}\ .$$
I shall only prove that $s_{n,d}\leq\ell({\bf p})$. Nevertheless I can't see how one could do better. 
In the ${\bf y}$-space ${\mathbb R}^d$ we consider a virtual box $$V:=[0,p_1]\times[0,p_2]\times\cdots\times[0,p_d]\subset{\mathbb R}^d\ .$$ This box is built up from  $d$-dimensional unit cubes, called chambers. The main diagonal $\sigma$ of this box connects ${\bf 0}$ with ${\bf p}=(p_1,\ldots ,p_d)$ and has length $\ell({\bf p})$. This diagonal passes a certain number of chambers in turn. We now interpret these chambers as $d$-dimensional faces of the original cube $C:=[0,1]^n\subset{\mathbb R}^n$, so that $\sigma$ can be viewed as an unfolding of an admissible path from ${\bf 0}$ to ${\bf 1}$ in $C$.
The intended "backfolding" of $\sigma$ is accomplished as follows: Assign each interval $[k-1,k]$ $(1\leq k\leq p_1)$ on the $y_1$-axis the number $k$, then assign each such interval on the $y_2$-axis one of the  next $p_2$ numbers, and so on along the other $y_i$-axes, until all numbers from $1$ to $n$ have so been distributed. We now identify each interval $0\leq x_k\leq 1$ on one of the axes of ${\mathbb R}^n$ with the corresponding $y_i$-interval on the proper axis of ${\mathbb R}^d$.
By projection onto the $y_i$-axes each piece $\Delta \sigma\subset \sigma$ lying in some chamber gets therewith  assigned $d$ different numbers from $[n]$, and these numbers indicate which coordinates $x_k$ shall be "active" along $\Delta \sigma$, in other words: on which $d$-dimensional face of $C$ the "backfolded" piece  $\Delta \sigma$ shall lie. It follows that $\sigma$ "backfolds" isometrically to an admissible path of length $\ell({\bf p})$ in $[0,1]^n$.
