By Sylvester's Law of Inertia, no orthonormal basis is possible, including the 2 by 2 case. In particular, your $H(e_{21},e_{21}) = 0.$
What is possible is to give a collection of matrices as below. There are $n$ matrices of type (A). There are $\frac{n^2 - n}{2}$ matrices of type (B).There are also $\frac{n^2 - n}{2}$ matrices of type (C).
Any two distinct matrices from the total collection of $n^2$ matrices are orthogonal. Given any matrix $\alpha$ of type (A), $H(\alpha, \alpha) = 1.$ Given any matrix $\beta$ of type (B), $H(\beta, \beta) = 1.$ However, given any matrix $\gamma$ of type (C), $H(\gamma, \gamma) = -1.$
(A) all your $e_{ii}$
(B) with all $j < k,$ take $ \frac{1}{\sqrt 2} (e_{jk} + e_{kj}) $
(C) with all $j < k,$ also take $ \frac{1}{\sqrt 2} (e_{jk} - e_{kj}) $
See SYLVESTER
EDIT, FRIDAY: this was a clever assignment, and not something I knew about. We have a quadratic form defined on square matices with real entries, given by $ q(M) = \mbox{tr}(M^2).$ This is positive definite on the linear subspace given by the symmetric matrices, as then $q(M^2) = \mbox{tr}(M M^T)$ which is the sum of the squares of all $n^2$ entries. It is negative definite on the skew-symmetric matrices, as then $q(M^2) = - \mbox{tr}(M M^T).$ By adding the dimensions, we find that the "corank" of the quadratic form $q$ is $0.$ Finally, any symmetric matrix $M$ and any skew symmetric matrix $N$ are orthogonal, as $ \mbox{tr}(M N) = \mbox{tr}(N^T M^T) = - \mbox{tr}(N M) = - \mbox{tr}(M N),$ so that $ \mbox{tr}(M N) =0$ here.
A symmetric matrix A can always be transformed in this way into a diagonal matrix D which has only entries 0, +1 and −1 along the diagonal. Sylvester's law of inertia states that the number of diagonal entries of each kind is an invariant of A, i.e. it does not depend on the matrix S used.
The number of +1s, denoted n+, is called the positive index of inertia of A, and the number of −1s, denoted n−, is called the negative index of inertia. The number of 0s, denoted n0, is the dimension of the kernel of A, and also the corank of A. These numbers satisfy an obvious relation
$n_0+n_{+}+n_{-}=n.$
The clever part is that we do not really need to talk about the "matrix" of $q,$ which would be an $n^2$ by $n^2$ matrix. To repeat, see QUADRATIC