If $G$ is a Cayley graph for $\mathbb{Z}_2^n$ with connection set $C \subseteq \mathbb{Z}_2^n \setminus \{0\}$, then for each element $a \in \mathbb{Z}_2^n$ there is an eigenvector $v$ given by
$$v_x = (-1)^{x \cdot a}$$
where $x \cdot a$ is the usual inner product when $x$ and $a$ are thought of as 01-vectors (i.e., it is the number of 1's they have in common). This eigenvector has eigenvalue
\begin{align*}\sum_{c \in C} (-1)^{c\cdot a} &= |\{c \in C: c \cdot a \equiv 0 \ \text{mod} \ 2 \}| - |\{c \in C: c \cdot a \equiv 1 \ \text{mod} \ 2 \}| \\
&= |C| - 2|\{c \in C: c \cdot a \equiv 1 \ \text{mod} \ 2 \}|
\end{align*}
This gives a full orthogonal set of eigenvectors for the graph $G$.

In your case, the graph can be described as a Cayley graph for $\mathbb{Z}_2^n$ with connection set consisting of all of the elements with $n/2$ 1's in them (when written as binary strings). We can think of the binary strings as subsets of the $n$ element set $[n]$. So for each subset $S \subseteq [n]$, we get an eigenvalue of
$$\sum_{T \subseteq [n], |T| = n/2} (-1)^{|S \cap T|}.$$
If $S$ has size $k$ then this is equal to
$$\sum_{i=0}^k \sum_{T \subseteq [n], |T| = n/2, |S \cap T| = i} (-1)^i = \sum_{i=0}^k (-1)^i\binom{k}{i}\binom{n-k}{n/2 - i}.$$
I may have stolen this from the aforementioned notes of @ChrisGodsil (I hope he does not mind), but it follows from the well-known technique for computing the eigenvalues of Cayley graphs for $\mathbb{Z}_2^n$ described above.

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