Asymptotic distribution of the sample variance: The sample variance does converges in distribution to the normal (under mild conditions), but there are better asymptotic approximations to the distribution, which respect the support of the sample variance. You can find relevant results and proof in O'Neill (2014) (Result 14, pp. 285, 293-4). So long as the kurtosis $\kappa$ of the underlying distribution is finite, you have:
$$\sqrt{n} \Bigg( \frac{S_n^2}{\sigma^2} - 1 \Bigg) \overset{\text{Dist}}{\longrightarrow} \text{N}(0, \kappa-1).$$
(The result is usually proved by writing out a decomposition of the sample variance and then using a combination of the classical central limit theorem and Slutsky's theorem.) For large $n$, this gives the large-sample approximating distribution:
$$S_n^2 \overset{\text{Approx}}{\sim} \text{N}(\sigma^2, n(\kappa-1) \sigma^4).$$
This approximation has the drawback that its support extends over negative values, which cannot be sample variance values. A better (asymptotically equivalent) aproximating distribution is:
$$S_n^2 \overset{\text{Approx}}{\sim} \sigma^2 \cdot \frac{\text{Chi-Sq}(DF_n)}{DF_n} \quad \quad \quad DF_n \equiv \frac{2n}{\kappa - (n-3)/(n-1)}.$$
This latter approximation respects the support of the sample variance, and it corresponds to the exact distribution of the sample variance in the special case where the underlying distribution is normal. For further discussion, see the linked paper.