The answer is $\frac{n(n-3)(n^2+6n-31)}{48}$.

For each $k$, let $d_k$ be the number of games that player $k$ wins. Note that
$$S := d_1 + d_2 + \dots + d_n = \binom{n}{2} - n = \frac{n(n-3)}{2}.$$
Observe that for a set of four players, if one of them beats everybody else, then the set is not harmonic. Also, there cannot be two players beating everybody else in the set simultaneously. Therefore, the number $N$ of harmonic sets is at most
$$\binom{n}{4} - \sum_{k=1}^{n} \binom{d_k}{3}.$$
Note that the binomial function $\binom{x}{3}$ is convex. So we can use the majorization inequality to get
$$N \le \binom{n}{4} - n \binom{\frac{n-3}{2}}{3} = \frac{n(n-3)(n^2+6n-31)}{48}.$$
It remains to show that this bound is attainable.

Suppose the game between player $k$ and player $k+1$ ends in a tie for all $k$, where player $n+1$ means player 1. For any other game between player $i$ and player $j$ where $1 \le i < j \le n$, player $j$ beats player $i$ if and only if $i \equiv j \pmod 2$.
For even $k$, player $k$ beats players $2, 4, \dots, k-2$ and $k+3, k+5, \dots, n$. So player $k$ has won $\frac{n-3}{2}$ games. For odd $k < n$, player $k$ beats players $1, 3, \dots, k-2$ and $k+3, k+5, \dots, n-1$. So player $k$ has won $\frac{n-3}{2}$ games. Also, player $n$ beats players $3, 5, \dots, n-2$. So player $n$ has won $\frac{n-3}{2}$ games. This shows all players have won the same number of games. Therefore, the equality of the majorization inequality holds.

We now show that for any set of four players in which nobody beats everyone else, it is harmonic. Once this is shown, all equalities in the above deduction hold, which means the bound is attained. Suppose on the contrary that there exists a non-harmonic set $S$ of four players $a, b, c, d$ such that none of them beats everyone else in this set.

* If there is no tie among the games played between $a, b, c, d$, then WLOG we may assume player $a$ beats players $b$ and $c$. By the assumption, player $d$ must beat player $a$. Since player $d$ does not beat everyone else, WLOG assume player $b$ beats player $d$. If player $c$ beats player $b$, then we have the cycle $a \to c \to b \to d \to a$, and so the set is harmonic. If player $c$ beats player $d$, WLOG assume player $b$ beats player $c$. Then we have the cycle $a \to b \to c \to d \to a$, and so the set is harmonic.

The only case left is that both players $b$ and $d$ beat player $c$, which means player $c$ is beaten by everyone else. This holds when each of $a, b, d$ is less than and has different parity as $c$, or is greater than and has the same parity as $c$. WLOG assume $a < b < d$.
- If $a < b < d < c$, then player $d$ beats everyone else, contradiction.
- If $a < b < c < d$, then player $b$ beats everyone else, contradiction.
- If $a < c < b < d$, then player $a$ beats everyone else, contradiction.
- If $c < a < b < d$, then player $d$ beats everyone else, contradiction.

* If there is a tie in the game played between $c$ and $d$, then $c$ and $d$ are consecutive integers, or are $1$ and $n$. By the construction, it is impossible that both $c$ and $d$ beat the same player, or are beaten by the same player. (This is the place where we have used $n$ is odd, since otherwise $1$ and $n$ are beaten by some players in the same construction of the even case.)

WLOG assume player $a$ does not beat player $c$, and player $d$ does not beat player $a$. If the same holds when $a$ is replaced by $b$, then we have a cycle $c \to a \to d \to c \to b \to d \to c$, and so the set is harmonic. (Although the cycle does not consist of distinct players, we can easily shorten the sequence so that it only consists of distinct players.) If player $b$ does not beat player $d$, and player $c$ does not beat player $b$, then we also have a cycle $c \to a \to d \to b \to c$, and so the set is harmonic.

This shows the construction works. So the proof is complete.