We can generalize domotorp's argument. If a harmonic set has $2k$ vectors in some $k$-dimensional space, then each basis in the set must contain exactly $k$ vectors, so every other vector must live in some complement to this subspace, so the harmonic set is reducible.
If we restrict our attention to irreducible harmonic sets, we then have at most $2k-1$ in each $k$-dimensional subspace. So to choose a basis for an $n$-dimensional irreducible harmonic set, we can choose any vector, then any other vector, then any vector other than the at most $3$ in the $2$-dimensional space generated by those $2$, then any vector other than the at most $5$ in the $3$-dimensional space generated by those $3$, and so on. This gives a lower bound for the number of bases at:
$$ \frac{ 2n (2n-1) (2n-3) \dots 1}{n (n-1) (n-2) \dots 1}$$
we also have the trivial upper bound of $\left( \begin{array}{c} 2n \\ n \end{array}\right)$. This gives us the possible intervals:
$n=1$: $[2,2]$
$n=2$: $[4,6]$
$n=3$: $[15,20]$
$n=4$: $[35,70]$
$n=5$: $[78.75,252]$
I think at some point, possibly $n=6$, these intervals start to overlap, which means we cannot rule out any further numbers. But we at least rule out all numbers which do not appear in these intervals: $8,10,12,14,22,$ and so on, from being the number of bases of an irreducible harmonic set. Since every harmonic set whose number of bases is not a multiple of $4$ is irreducible, we can rule out $10$, $14$, $22$, $26$, $30$, $34$, $74$, and $78$ as being the number of bases of a harmonic set.
$\left( \begin{array}{c} 2n \\ n \end{array}\right)$ is achievable for each $n$ by $2n$ generic vectors in $n$-dimensional space. This means we can get $6$ and $12=2 \times 6$, so the first number I don't know if we can get is $18$.