The question is a little bit too general in its present form to get a useful result. Nevertheless, with some slight restrictions we can get a useful general form for the asymptotic distribution using the delta method. To do this, let's assume that the underlying distribution for the data has a finite mean $\mu$ and finite variance $\sigma^2$. This allows us to apply the central limit theorem to get the asymptotic distribution $\bar{X}_n \sim \text{N}(\mu, \sigma^2/n)$, which means that $\bar{X}_n \rightarrow \mu$ as $n \rightarrow \infty$. Since the sample mean gets closer and closer to the true mean in the limit, we can see that the asymptotic distribution of $g(\bar{X}_n)$ will be fully determined by the local behaviour of the function $g$ in a neighbourhood of the point $\mu$.
To proceed further, let's make some mild assumptions about this local behaviour. Specifically, we will reduce the scope of the allowable functions by assuming that $g$ is an analytic function at $\mu$ (i.e., the function $g$ is infinitely differentiable at this point and it is representable by its Taylor series representation at this point). Thus, for all points $x$ in a neighbourhood of $\mu$ we can write the function $g$ as:
$$g(x) = g(\mu) + \sum_{k=1}^\infty \frac{g^{(k)}(\mu)}{k!} \cdot (x-\mu)^k.$$
The asymptotic distribution depends on which derivatives of the function are zero at the point $\mu$. Without loss of generality, let $K = \min \{ k = 1,2,3,... | g^{(k)}(\mu) \neq 0 \}$ denote the order of the first non-zero derivative of the function at the mean of the random variable under analysis, which means we have $g^{(1)}(\mu) = \cdots = g^{(K-1)}(\mu) = 0$. To facilitate our analysis we also define the standardised sample mean $Z_n \equiv \sqrt{n} (\bar{X}_n-\mu)/\sigma$. We can now write the Taylor expansion of interest as:
$$\begin{aligned}
g(\bar{X}_n)
&= g(\mu) + \sum_{k=K}^\infty \frac{g^{(k)}(\mu)}{k!} \cdot (\bar{X}_n-\mu)^k \\[6pt]
&= g(\mu) + \sum_{k=K}^\infty \frac{g^{(k)}(\mu)}{k!} \cdot \sigma^k \cdot \frac{Z_n^k}{n^{k/2}}. \\[6pt]
\end{aligned}$$
As we take $n \rightarrow \infty$ we can apply the classical central limit theorem to get the asymptotic distribution $Z_n \sim \text{N}(0,1)$, and so the higher order terms in this expansion will converge to zero much more rapidly than the term of order $K$. The asymptotic distribution of our function at the sample mean will be determined by the $K$th term of the Taylor series:
$$\begin{aligned}
g(\bar{X}_n)
&\sim g(\mu) + \frac{g^{(K)}(\mu)}{K!} \cdot \sigma^K \cdot \frac{Z_n^K}{n^{K/2}}. \\[6pt]
\end{aligned}$$
We can see from this result that the asymptotic distribution is heavily dependent on the order value $K$. If $K=1$ then the asymptotic distribution will be a normal distribution, if $K=2$ then the asymptotic distribution will be a scaled chi-squared distribution, if $K=3$ then the asymptotic distribution is for the scaled version of the cube of a normal random variable (see here for discussion), and so on.
