Let $Q_n$ be the hypercube graph formed from the vertices and edges of an $n$-dimensional hypercube and $D_m(G)$ be the distortion of graph $G$ in Euclidean space $\mathbb{R}^m$.
- Is it true that $D_m(Q_n) \leq D_{m-1}(Q_n)$?
- Is it true for arbitrary graph?
An embedding of $G$ in $\mathbb{R}^m$ is a map $f$ associating the vertices of $G$ with points in $\mathbb{R}^m$.
Denote by
$E_m(G)$ the set of all embeddings of $G$ in $\mathbb{R}^m$,
$d_G(u, v)$ the number of edges in a shortest path from $u$ to $v$ ($u, v \in V(G)$).
$$distortion_m(G) = \min_{f \in E_m(G)} \frac{ \max_{u, v \in G}\frac{ ||f(u) - f(v)||}{d_G(u, v)}}{\min_{u, v \in G}\frac{||f(u) - f(v)||}{d_G(u, v)}}$$
For any fixed graph $G$ with $\vert V(G)\vert=n$, and for any $m$, $D_m(G)\leq D_{m-1}(G)$, as any embedding into $\mathbb{R}^{m-1}$ can be trivially viewed inside of $\mathbb{R}^m$ by just adding a zero coordinate to any such mapping. This doesn't change Euclidean distances, so attains the same distortion. This means the left hand side is an infimum over a larger set, so is at most the right hand side. Of course, this holds for $G=Q_n$.
For the specific case of the hypercube, you may be interested in the well-known result that $D_n(Q_n)=\sqrt{n}$. In fact, $\inf_m D_m(Q_n)=D_n(Q_n)=\sqrt{n}$, so the best distortion can be realized via an embedding into dimension $n$. The exact embedding maps $Q_n$ bijectively into $\{0,\sqrt{n}\}^n$ in the obvious way. The graph distance between vertices of $Q_n$ is just the Hamming distance, and it is easy to see that if $\Delta_{Q_n}(u,v)=k$, then $\|F(u)-F(v)\|_2=\sqrt{kn}$, so the ratio is $\sqrt{\frac{n}{k}}$. This is at least $1$ with equality on $(0,\ldots,0)$ and $(\sqrt{n},\ldots,\sqrt{n})$, and at most $\sqrt{n}$ on adjacent vertices. To prove that this is optimal, even when going to higher dimensions, one can use Boolean analysis: see for instance Exercise 2.53 of O'Donnell's excellent book "Analysis of Boolean Functions" (freely available online).