What are the skewness and kurtosis of the sample mean? Consider a sequence of IID random variables $X_1,X_2,X_3,...$ from a common distribution with mean $\mu$, variance $\sigma^2$, skewness $\gamma$ and kurtosis $\kappa$ (all finite).  The mean and variance of the sample mean are well-known:
$$\mathbb{E}(\bar{X}_n) = \mu
\quad \quad \quad \quad \quad 
\mathbb{V}(\bar{X}_n) = \frac{\sigma^2}{n}.$$
What are the corresponding formulae for the skewness and kurtosis of the sample mean?

$$\begin{align}
\boxed{
\quad \quad \ \ \ \mathbb{E}(\bar{X}_n) = \mu
\quad \quad \quad \quad \quad \quad \quad \ \
\mathbb{V}(\bar{X}_n) = \frac{\sigma^2}{n}, \\[12pt]
\quad \mathbb{Skew}(\bar{X}_n) = \frac{\gamma}{\sqrt{n}}
\quad \quad \quad \quad \quad 
\mathbb{Kurt}(\bar{X}_n) = 3 + \frac{\kappa - 3}{n}. \quad \\}
\end{align}$$
The mean, variance, skewness and kurtosis of the sample mean are shown in the box above.  These formulae are valid for any case where the underlying values are IID with finite kurtosis.  It is simple to confirm that $\mathbb{Skew}(\bar{X}_n) \rightarrow 0$ and $\mathbb{Kurt}(\bar{X}_n) \rightarrow 3$ as $n \rightarrow \infty$, which means that the sample mean is asymptotically unskewed and mesokurtic.  This is also implied by the classical central limit theorem, which ensures that the standardised sample mean converges in distribution to the normal distribution.

Proof via cumulants: The simplest way to prove these results is via moment/cumulant generating functions.  Suppose we let $m_X$ and $K_X$ denote the moment generating function and cumulant generating function for the underlying IID random variables in the sequence.  It is simple to show that $m_{\bar{X}_n}(t) = m_X(t/n)^n$ so the cumulant generating function for the sample mean has the form:
$$K_{\bar{X}_n}(t) = n K_X(t/n).$$
Consequently, the cumulants of the sample mean are related to the cumulants of the underlying random variables by:
$$\begin{align}
\bar{\kappa}_r 
\equiv \frac{d^r}{dt^r} K_{\bar{X}_n} (t) \Bigg|_{t=0} 
&= n \frac{d^r }{dt^r} K_X(t/n) \Bigg|_{t=0} \\[6pt]
&= \frac{1}{n^{r-1}} K_X^{(r)}(t/n) \Bigg|_{t=0} \\[6pt]
&= \frac{1}{n^{r-1}} K_X^{(r)}(0) \\[6pt]
&= \frac{\kappa_r}{n^{r-1}}. \\[6pt]
\end{align}$$
Using the relationship of the cumulants to the moments of interest, we then have:
$$\begin{align}
\mathbb{V}(\bar{X}_n)
&= \bar{\kappa}_2 \\[12pt]
&= \frac{\kappa_2}{n} \\[6pt]
&= \frac{\sigma^2}{n}, \\[6pt]
\mathbb{Skew}(\bar{X}_n)
&= \frac{\bar{\kappa}_3}{\bar{\kappa}_2^{3/2}} \\[6pt]
&= \frac{\kappa_3 / n^2}{(\kappa_2/n)^{3/2}} \\[6pt]
&= \frac{1}{\sqrt{n}} \cdot \frac{\kappa_3}{\kappa_2^{3/2}} \\[6pt]
&= \frac{\gamma}{\sqrt{n}}, \\[6pt]
\mathbb{Kurt}(\bar{X}_n)
&= \frac{\bar{\kappa}_4 + 3 \bar{\kappa}_2^2}{\bar{\kappa}_2^2} \\[6pt]
&= \frac{\kappa_4/n^3 + 3 (\kappa_2/n)^2}{(\kappa_2/n)^2} \\[6pt]
&= \frac{\kappa_4/n + 3 \kappa_2^2}{\kappa_2^2} \\[6pt]
&= \frac{(\kappa \sigma^4 - 3\sigma^4)/n + 3 \sigma^4}{\sigma^4} \\[6pt]
&= 3 + \frac{\kappa - 3}{n}. \\[6pt]
\end{align}$$
This method gives the results of interest, and it can also be generalise to give corresponding results for higher-order moments.  As can be seen, the higher-order results are particularly simple, but the corresponding relationships for higher-order moments get messy once you get to high order.

Proof via expansion to raw moments: An alternative method of deriving these results is to expand the relevant central moments for the sample mean and simplify down to raw moments of the underlying random variables.  Let $Y_i \equiv X_i - \mu$ and note that these random variables have mean zero, but have the same higher-order moments as $X_i$.  The relevant higher-order central moments for the sample mean are:$^\dagger$
$$\begin{align}
\mathbb{E}((\bar{X}_n - \mu)^3)
&= \mathbb{E} \Bigg( \bigg( \frac{1}{n} \sum_{i=1}^n (X_i - \mu) \bigg)^3 \Bigg) \\[6pt]
&= \mathbb{E} \Bigg( \bigg( \frac{1}{n} \sum_{i=1}^n Y_i \bigg)^3 \Bigg) \\[6pt]
&= \frac{1}{n^3} \cdot \mathbb{E} \Bigg( \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n Y_i Y_j Y_k \Bigg) \\[8pt]
&= \frac{1}{n^3} \cdot \mathbb{E} \Bigg( \sum_{i} Y_i^3 + \sum_{i \neq j} Y_i^2 Y_j + \sum_{i \neq j \neq k} Y_i Y_j Y_k \Bigg) \\[8pt]
&= \frac{1}{n^3} \cdot \sum_{i} \mathbb{E}(Y_i^3) \\[12pt]
&= \frac{1}{n^3} \cdot \sum_{i} \gamma \sigma^3 \\[12pt]
&= \frac{1}{n^3} \cdot n \gamma \sigma^3 \\[12pt]
&= \frac{\gamma}{n^2} \cdot \sigma^3, \\[12pt]
\mathbb{E}((\bar{X}_n - \mu)^4)
&= \mathbb{E} \Bigg( \bigg( \frac{1}{n} \sum_{i=1}^n (X_i - \mu) \bigg)^4 \Bigg) \\[6pt]
&= \mathbb{E} \Bigg( \bigg( \frac{1}{n} \sum_{i=1}^n Y_i \bigg)^4 \Bigg) \\[6pt]
&= \frac{1}{n^4} \cdot \mathbb{E} \Bigg( \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n \sum_{l=1}^n Y_i Y_j Y_k Y_l \Bigg) \\[6pt]
&= \frac{1}{n^4} \cdot \mathbb{E} \Bigg( \sum_{i} Y_i^4 + \sum_{i \neq j} Y_i^3 Y_j + \sum_{i \neq j} Y_i^2 Y_j^2 \\[12pt]
&\quad \quad \quad \quad \quad \quad + \sum_{i \neq j \neq k} Y_i^2 Y_j Y_k + \sum_{i \neq j \neq k \neq l} Y_i Y_j Y_k Y_l \Bigg) \\[6pt]
&= \frac{1}{n^4} \cdot \Bigg[ \sum_{i} \mathbb{E}(Y_i^4) + \sum_{i \neq j} \mathbb{E}(Y_i^2) \mathbb{E}(Y_j^2) \Bigg] \\[6pt]
&= \frac{1}{n^4} \cdot \Bigg[ \sum_{i} (\kappa \sigma^4) + \sum_{i \neq j} \sigma^4 \Bigg] \\[6pt]
&= \frac{1}{n^4} \cdot \Bigg[ n \kappa \sigma^4 + 3n(n-1) \sigma^4 \Bigg] \\[6pt]
&= \frac{(\kappa + 3(n-1)) \sigma^4}{n^3} \\[6pt]
&= \frac{3n + (\kappa - 3)}{n^3} \cdot \sigma^4. \\[6pt]
\end{align}$$
Consequently, the skewness and kurtosis of the sample mean are given respectively by:
$$\begin{align}
\mathbb{Skew}(\bar{X}_n)
&= \frac{\mathbb{E}((\bar{X}_n - \mu)^3)}{\mathbb{V}(\bar{X}_n)^{3/2}} \\[6pt]
&= \frac{\gamma}{n^2} \cdot \sigma^3 \bigg/ \frac{\sigma^3}{n^{3/2}} \\[6pt]
&= \frac{\gamma}{\sqrt{n}}, \\[12pt]
\mathbb{Kurt}(\bar{X}_n)
&= \frac{\mathbb{E}((\bar{X}_n - \mu)^4)}{\mathbb{V}(\bar{X}_n)^2} \\[6pt]
&= \frac{3n + (\kappa - 3)}{n^3} \cdot \sigma^4 \bigg/\frac{\sigma^4}{n^2}
\quad \quad \quad \quad \quad \quad \quad  \\[6pt]
&= \frac{3n + (\kappa - 3)}{n} \\[12pt]
&= 3 + \frac{\kappa - 3}{n}. \\[12pt]
\end{align}$$

$^\dagger$ We have used a slight abuse of notation in the range of the summations; when we refer to e.g., $i \neq j \neq k$ we use this as shorthand for the set of indices where all the indices are distinct.  That is, we do not read these inequalities in their strict meaning, but read them as if the inequality were intended to be transitive.
