There are certainly special graphs that are always Hamiltonian (if every vertex of a graph of n vertices has degree at least n/2, say) and these have efficient algorithms associated with them.
For instance, this paper proves the graph of a random 5-outregular digraph is Hamiltonian and there is an algorithm that finds a Hamiltonian cycle in polynomial time.
Your sequence is Sloane A096368. Sloane links to this page, which has files of all examples up to $13$ vertices. MathSciNet has 30 papers with "regular tournament" in the title, none of which seem to know much about enumeration up to isomorphism. A quick scan of the papers with "balanced tournament" in the title suggests that that term means something else, so I would search on "regular tournament".
McKay proves an asymptotic formula for the number of LABELED regular tournaments of the form $2^{\binom{n}{2}} e^{-O(n \log n)}$. (See his paper for a much more precise statement.) Since the size of $n!$ is "only" $e^{O(n \log n)}$, we can deduce that the number of isomorphism classes is also $2^{\binom{n}{2}} e^{-O(n \log n)}$, and in particular goes to $\infty$ as $n \to \infty$.
I can say a bit more about the labeled problem, where we don't quotient by automorphism. This is Sloane A007079.
The number we want is the coefficient of $\prod_{i=1}^{n} x_i^{(n-1)/2}$ in $\prod_{1 \leq i < j \leq n} (x_i+x_j)$; each factor corresponds to an edge and choosing $x_i$ or $x_j$ corresponds to orienting this edge. This polynomial is the Schur polynomial $s_{(n-1)(n-2)\cdots 321}(x_1, \ldots, x_n)$; this follows from the ratio of alternants formula. So you are looking for the Kotska number
$$K_{(n-1)(n-2)\cdots 321,\ mmm \cdots m} \ \mbox{where} \ m=(n-1)/2.$$
I don't think there is a closed formula for this Kotska number.
Here's something interesting that I couldn't get to work, but maybe will help someone else. Essentially by definition, this Kostka number is the dimension of the space of diagonal invariants in the representation of $SL_n$ associated to the partition $(n-1, n-2, \cdots, 2,1)$. And this vector space comes with a natural $S_n$-action, from the embedding of $S_n$ into $SL_n$.
Unfortunately, they don't match! When $n=3$, there are two labeled tournaments. (Orient the triangle clockwise or counterclockwise.) So a permutation $\sigma$ in $S_3$ either preserves or switches the tournaments based on the sign of $\sigma$. The corresponding permutation representation of $S_3$ is the direct sum of the trivial and the sign rep. By way of contrast, the representation of $S_3$ on the diagonal invariants of $V_{21}$ is the $2$-dimensional irrep of $S_3$.
Still, it would be really cool if we could find a deeper relation between these two representations of $S_n$ than simply the fact that they have the same dimension. In particular, remember that the number of orbits in a permutation representation is always the multiplicity of the trivial rep, so one could imagine counting isomorphism classes by representation theory if we can find a good alternate description of the tournament representation.
Best Answer
Only a longer comment, not a complete answer.
I think of a $2n \times 2n$ chess board, where paths connect midpoints of squares. Let the total area of the board be $2n \times 2n$. Then the closed region (the tree), encloses an area of independent of the cycle, namely $2n^2-1$.
I have not proved this, but it seems straightforward to prove that trees with certain properties, and Hamiltonian cycles are in bijection.
When $n$ is odd, we can make a rotationally-symmetric tree, (where each of the four arm is a spiral, for example) and thus constructing a balanced path, since there is center vertex for the tree, so the $2n^2-2$ (which is divisible by $4$), can be distributed evenly in the four quadrants.
If $n$ is even, we still can put one vertex of the tree in the middle, but the $2n^2-2$ is now not divisible by $4$, and I believe this can be exploited to somehow show that said circuit is impossible.