Variance of sample variance? What is the variance of the sample variance? In other words I am looking for $\mathrm{Var}(S^2)$.

I have started by expanding out $\mathrm{Var}(S^2)$ into $E(S^4) - [E(S^2)]^2$

I know that $[E(S^2)]^2$ is $\sigma$ to the power of 4. And that is as far as I got.
Here's a general derivation that does not assume normality. 

Let's rewrite  the sample variance $S^2$  as an average over all pairs of indices:
$$S^2={1\over{n\choose 2}}\sum_{\{i,j\}} {1\over2}(X_i-X_j)^2.$$ 
Since  $\mathbb{E}[(X_i-X_j)^2/2]=\sigma^2$, we see that $S^2$ is an unbiased estimator for $\sigma^2$. 

The variance of $S^2$ is the expected value of 
$$\left({1\over{n\choose 2}}\sum_{\{i,j\}} \left[{1\over2}(X_i-X_j)^2-\sigma^2\right]\right)^2.$$   

When you expand the outer square, there are 3 types of cross product terms 
 $$\left[{1\over2}(X_i-X_j)^2-\sigma^2\right] \left[{1\over2}(X_k-X_\ell)^2-\sigma^2\right]$$
depending on the size of the intersection $\{i,j\}\cap\{k,\ell\}$. 


When this intersection is empty, the factors are independent and the expected cross product is zero. 
There are $n(n-1)(n-2)$ terms where  $|\{i,j\}\cap\{k,\ell\}|=1$ and each has an expected cross product of $(\mu_4-\sigma^4)/4$. 
There are ${n\choose 2}$ terms where  $|\{i,j\}\cap\{k,\ell\}|=2$ and each has an expected cross product of $(\mu_4+\sigma^4)/2$. 


Putting it all together shows that $$\mbox{Var}(S^2)={\mu_4\over n}-{\sigma^4\,(n-3)\over n\,(n-1)}.$$ Here $\mu_4=\mathbb{E}[(X-\mu)^4]$ is the fourth central moment of $X$.
I will use this as an example of this theorem(from Seber, G.A. and Lee, A.J. (2012))
Let $X_1, X_2, ... , X_n$ be independent rvs with means $(\theta_1, \theta_2, ... ,\theta_n)$,common $\mu_2,\mu_3,\mu_4$. If A is any n x n symmetric matrix and $a$ is a column vector of the diagonal elements of A, then
$$var[X'AX]=(\mu_4-3\mu^2_2)a'a+2\mu^2_2tr(A^2)+4\mu_2\theta'A^2\theta+4\mu_3\theta'Aa
$$
denote $1_n$ as n-dim column vector that all elements are 1, notice that for sample variance
$$S^2=\frac{1}{n-1}X'AX, where A=I_n-\frac{1}{n}1_n1_n'
$$
and we have$A^2=A$, $a=(1-\frac{1}{n})1_n$
since $X_i$ in our case are iid, let's say their mean is $\mu$, then $\theta=\mu1_n$
so the third and fourth term is $0$,since
$$
A^2\theta=A\theta=\mu(1_n-\frac{1}{n}1_n(1_n'1_n))=0\\
Aa=(1-\frac{1}{n})(1_n-\frac{1}{n}1_n(1_n'1_n))=0
$$
then$$
var[S^2]=\frac{1}{(n-1)^2}[(\mu_4-3\mu_2^2)(1-\frac{1}{n})^2n+2\mu_2^2(n-1)]=\frac{\mu_4}{n}-\frac{n-3}{n(n-1)}\mu_2^2
$$