$\def\FF{\mathbb{F}}\def\PP{\mathbb{P}}$I can achieve the bound $\binom{n-1}{2}$ if $n=q+1$ for a prime power $q$.

**Outline of method:** Suppose that $X$ is a set of size $n$ and $G$ is a group acting on $X$ in a sharply three transitive manner. This means that, is $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ are two ordered triples of distinct elements of $G$, there is precisely one $g \in G$ with $g(x_i)=y_i$ for $i=1$, $2$, $3$. (So, in particular, we have $|G| = n(n-1)(n-2)$.) Suppose furthermore that there is an element $\gamma$ in $G$ which acts by an $n$-cycle on $X$, and $\gamma$ is conjugate to $\gamma^{-1}$.

Then I claim $n$ is achievable. Let $C \subset G$ be the conjugacy class of $\gamma$. We note that the only permutations in $S_n$ which commute with an $n$-cycle are powers of that $n$-cycle, so $Z(\gamma) = \langle \gamma \rangle$ and the size of the conjugacy class $C$ is $|G|/|Z(\gamma)| = n(n-1)(n-2)/n=(n-1)(n-2)$.

I claim that, for any distinct elements $y_1$, $y_2$, $y_3$ in $X$, there is a unique $\delta \in C$ with $\delta(y_1) = y_2$ and $\delta(y_2) = y_3$. We first see that there is at least one $\delta$: Fix $x_1 \in X$, define $x_2 = \gamma(x_1)$ and $x_3 = \gamma(x_2)$, and let $h \in G$ be such that $h(x_i) = y_i$. Then $\delta = h \gamma h^{-1}$ does the job. But $|C| = (n-1) (n-2)$ and each $\delta \in C$ gives rise to $n$ triples $(y_1, y_2, y_3)$ with $\delta(y_1) = y_2$, $\delta(y_2) = y_3$. So, if each triple occurs at most once, then each triple occurs exactly once.

We have shown that the elements of $C$ are $(n-1)(n-2)$ oriented $n$-cycles, with each $y_1 \to y_2 \to y_3$ occurring in exactly one. But also, we assumed $\gamma$ conjugate to $\gamma^{-1}$, so for every cycle in $C$ its reverse is also in $C$. Identifying these, we get $\binom{n-1}{2}$ unoriented cycles, where for each $y_2$, and each pair of neighbors $y_1$, $y_3$, there is exactly one cycle where they occur.

**Details** Take $X = \PP^1(\FF_q)$ and $G = PGL_2(\FF_q)$ with the obvious action. It is well known that $G$ is sharply triply transitive. Choose an identification of $(\FF_q)^2$ with the field $\FF_{q^2}$, choose a generator $\theta$ for the cyclic group $\FF_{q^2}^{\times}$ and let $\gamma \in GL_2(\FF_q)$ be the matrix of multiplication by $\theta$. Then the image of $\gamma$ in $PGL_2(\FF_q)$ has order $q+1$. Moreover, in $PGL_2(\FF_q)$, every element is conjugate to its inverse. This proves the claim. $\square$.

Zassenhaus classified the sharply triple transitive permutation groups and, for all of them, $|X|=q+1$ for a prime power $q$. So there aren't any other $n$ achievable in this way.