Asymptotic distribution of sample variance of non-normal sample This is a more general treatment of the issue posed by  this question.
After deriving the asymptotic distribution of the sample variance, we can apply the Delta method to arrive at the corresponding distribution for the standard deviation.  
Let a sample of size $n$ of i.i.d. non-normal random variables $\{X_i\},\;\; i=1,...,n$, with mean $\mu$ and variance $\sigma^2$. Set the sample mean and the sample variance as 
$$\bar x = \frac 1n \sum_{i=1}^nX_i,\;\;\; s^2 = \frac 1{n-1} \sum_{i=1}^n(X_i-\bar x)^2$$
We know that 
$$E(s^2) = \sigma^2, \;\;\; \operatorname {Var}(s^2) = \frac{1}{n} \left(\mu_4 - \frac{n-3}{n-1}\sigma^4\right)$$
where $\mu_4 = E(X_i -\mu)^4$, and we restrict our attention to distributions for which what moments need to exist and be finite, do exist and are finite.
Does it hold that
$$\sqrt n(s^2 - \sigma^2) \rightarrow_d N\left(0,\mu_4 - \sigma^4\right)\;\; ?$$
 A: The excellent answers by Alecos and JohnK already derive the result you are after, but I would like to note something else about the asymptotic distribution of the sample variance.
It is common to see asymptotic results presented using the normal distribution, and this is useful for stating the theorems.  However, practically speaking, the purpose of an asymptotic distribution for a sample statistic is that it allows you to obtain an approximate distribution when $n$ is large.  There are lots of choices you could make for your large-sample approximation, since many distributions have the same asymptotic form.  In the case of the sample variance, it is my view that an excellent approximating distribution for large $n$ is given by:
$$\frac{S_n^2}{\sigma^2} \sim \frac{\text{Chi-Sq}(\text{df} = DF_n)}{DF_n},$$
where $DF_n \equiv 2 / \mathbb{V}(S_n^2 / \sigma^2) = 2n / ( \kappa - (n-3)/(n-1))$ and $\kappa = \mu_4 / \sigma^4$ is the kurtosis parameter.  This distribution is asymptotically equivalent to the normal approximation derived from the theorem (the chi-squared distribution converges to normal as the degrees-of-freedom tends to infinity).  Despite this equivalence, this approximation has various other properties you would like your approximating distribution to have:

*

*Unlike the normal approximation derived directly from the theorem, this distribution has the correct support for the statistic of interest.  The sample variance is non-negative, and this distribution has non-negative support.


*In the case where the underlying values are normally distributed, this approximation is actually the exact sampling distribution.  (In this case we have $\kappa = 3$ which gives $DF_n = n-1$, which is the standard form used in most texts.)  It therefore constitutes a result that is exact in an important special case, while still being a reasonable approximation in more general cases.

Derivation of the above result: Approximate distributional results for the sample mean and variance are discussed at length in O'Neill (2014), and this paper provides derivations of many results, including the present approximating distribution.
This derivation starts from the limiting result in the question:
$$\sqrt{n} (S_n^2 - \sigma^2) \sim \text{N}(0, \sigma^4 (\kappa - 1)).$$
Re-arranging this result we obtain the approximation:
$$\frac{S_n^2}{\sigma^2} \sim \text{N} \Big( 1, \frac{\kappa - 1}{n} \Big).$$
Since the chi-squared distribution is asymptotically normal, as $DF \rightarrow \infty$ we have:
$$\frac{\text{Chi-Sq}(DF)}{DF} \rightarrow \frac{1}{DF}
\text{N} ( DF, 2DF ) = \text{N} \Big( 1, \frac{2}{DF} \Big).$$
Taking $DF_n \equiv 2 / \mathbb{V}(S_n^2 / \sigma^2)$ (which yields the above formula) gives $DF_n \rightarrow 2n / (\kappa - 1)$ which ensures that the chi-squared distribution is asymptotically equivalent to the normal approximation from the limiting theorem.
A: To side-step dependencies arising when we consider the sample variance, we write 
$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$
$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$
and after a little manipualtion,
$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$
Therefore
$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2-  \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2  $$
Manipulating,
$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2-  \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2  $$
$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2-  \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$
$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$
The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$. 
We also have $\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$. The first component converges in distribution to a Normal, the second convergres in probability to zero. Then by Slutsky's theorem the product converges in probability to zero,
$$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$
We are left with the term
$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$
Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have
$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$
and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and
$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$
Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.
A: You already have a detailed answer to your question but let me offer another one to go with it. Actually, a shorter proof is possible based on the fact that the distribution of 
$$S^2 = \frac{1}{n-1} \sum_{i=1}^n \left(X_i -  \bar{X} \right)^2 $$
does not depend on $E(X) = \xi$, say. Asymptotically, it also does not matter whether we change the factor $\frac{1}{n-1}$ to $\frac{1}{n}$, which I will do for convenience. We then have
$$\sqrt{n} \left(S^2 - \sigma^2 \right) = \sqrt{n} \left[ \frac{1}{n} \sum_{i=1}^n X_i^2 - \bar{X}^2 -  \sigma^2 \right]$$
And now we assume without loss of generality that $\xi = 0$ and we notice that
$$ \sqrt{n} \bar{X}^2 = \frac{1}{\sqrt{n}} \left( \sqrt{n} \bar{X} \right)^2$$
has probability limit zero, since the second term is bounded in probability (by the CLT and the continuous mapping theorem), i.e. it is $O_p(1)$. The asymptotic result now follows from Slutzky's theorem and the CLT, since
$$\sqrt{n} \left[ \frac{1}{n} \sum X_i^2 - \sigma^2 \right] \xrightarrow{D} \mathcal{N} \left(0, \tau^2 \right)$$
where $\tau^2 = Var \left\{ X^2\right\} = \mathbb{E} \left(X^4 \right) - \left( \mathbb{E} \left(X^2\right) \right)^2$. And that will do it.
