Concerning the edges, consider the following:

Let $v\in \{0,1\}^d$ be a 0/1-vector with exactly $k$ zeroes. Then there exist $2^k$ 0/1-vectors that have vanishing inner product with $v$. The number of non-edges then is

$$\frac 12 \sum_{k=0}^d {d\choose k} 2^k = \frac {3^d}2.$$

Concerning the edges, consider the following:

Let $v\in \smash{\{0,1\}^d}$ be a 0/1-vector with exactly $k$ zeroes. If $v\not=0$, there are exactly $2^k$ 0/1-vectors that have vanishing inner product with $v$. The vector $v=0$ is an exception, as it has zero inner product with itself (this would give a loop otherwise), and so we count only $\smash{2^d-1}$ neighbors for this one. The number of non-edges then is

$$\frac 12 \Big[\sum_{k=0}^d {d\choose k} 2^k-1\Big] = \frac {3^d-1}2.$$

Concerning the edges, consider the following:

Let $v\in \{0,1\}^d$ be a 0/1-vector with exactly $k$ zeroes. Then there exist $2^k$ 0/1-vectors that have vanishing inner product with $v$. The number of edges then is

$$\frac 12 \sum_{k=0}^d {d\choose k} 2^k = \frac {3^d}2.$$

Concerning the edges, consider the following:

Let $v\in \{0,1\}^d$ be a 0/1-vector with exactly $k$ zeroes. Then there exist $2^k$ 0/1-vectors that have vanishing inner product with $v$. The number of non-edges then is

$$\frac 12 \sum_{k=0}^d {d\choose k} 2^k = \frac {3^d}2.$$