It seems to me that $H_{N}$ is the character table of an elementary Abelian $2$-group of order $2^{n}$ (with respect to a suitable ordering of elements). As such, its rows are orthogonal by the orthogonality relations for group characters. Also, it is clear, by induction that $H_{N}$ is symmetric.Hence we have $H_{N}H_{N}^{t} = 2^{n}I$ (since it is a character table) and $H_{N}^{2} = 2^{n}I$. Thus the eigenvalues of $H_{N}$ are $\pm \sqrt{2^{n}}$, as already noted by Carlos Beenakker.

Note also that $H_{N}$ has trace zero for $N > 1,$ so that both square roots occur with equal multiplicity as eigenvalues.

Note that since $H_{N}$ is a character table of an Abelian group for $N \geq 2$, its rows and columns are mutually orthogonal. Now since $T = \frac{H_{N}}{2^{\frac{n}{2}}}$ is a matrix of multiplicative order two, we have $T\frac{I+T}{2} = \frac{I+T}{2}$ and likewise $T\frac{I-T}{2} = -\frac{I-T}{2}.$ Hence the columns of $\frac{I+T}{2}$ are eigenvectors of $T$ with eigenvalue $1$ and the columns of $\frac{I-T}{2}$ are eigenvectors of $T$ with eigenvalue $-1$. We can also see that $\frac{I+T}{2}$ and $\frac{I-T}{2}$ are mutually orthogonal idempotent matrices with sum $I$.

It follows that $\frac{I+T}{2}$ has rank $2^{n-1}$, as does $\frac{I-T}{2}.$ Hence the columns of $\frac{I}{2} + \frac{H_{N}}{2^{1+\frac{n}{2}}}$ are eigenvectors of $H$ with eigenvalue $2^{\frac{n}{2}}$ spanning the $2^{\frac{n}{2}}$-eigenspace and the columns of $\frac{I}{2} - \frac{H_{N}}{2^{1+\frac{n}{2}}}$ are eigenvectors of $H$ with eigenvalue $-2^{\frac{n}{2}}$ spanning the
$-2^{\frac{n}{2}}$-eigenspace.