$G$ above corresponds to an $n$-dimensional hypercube, so $G'=(V',E')$ is necessarily a subgraph of the hypercube. Let $v$ be any vertex in $V'$ (and hence also in $V$). Note that in a hypercube the number of vertices a distance $D$ in the $\ell^1$ norm from $v_0$ has exactly $D$ edges which lead to vertices a distance $D-1$ from $v$ and $n - D$ vertices a distance $D+1$ from $v$. Since $v \in V'$, at least $d$ vertices at distance $1$ from $v$ must be in $V'$. At distance 2 there are $d(d-1)$ incoming edges, but each site has only $2$ edges which connect to sites a distance $1$ from $v$, and hence there must be at least $d(d-1)/2 = \binom{d}{2}$ vertices a distance 2 from $v$. Now assume there are at least $\binom{d}{D}$ vertices in $V'$ that are a distance $D$ from $v$. Then there are $\binom{d}{D}(d-D)$ edges connecting these vertices to vertices $D+1$ from $v$. However each of these has at most $D+1$ edges connecting to vertices $D$ from $v$, and hence there is at least $\binom{d}{D}\frac{d-D}{D+1} = \binom{d}{D+1}$ vertices at distance $D+1$. Thus, by induction, there are at least $\sum_{D=0}^{d} \binom{d}{D} = 2^d$ vertices in $V'$.

A trivial example to show this bound is tight is to take a hypercube of dimension $d$ on the boundary of the original hypercube, as this has exactly $2^d$ vertices and has degree $d$ for all vertices.