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.

This is quite a well-known result in statistics, and it can be found in a number of books and papers on sampling theory.  You can find a range of useful moment results of this kind in O'Neill (2014) (this one is given in Result 3, p. 284).  Consider a distribution with mean $\mu$, variance $\sigma^2$, skewness $\gamma$ and kurtosis $\kappa$ (where all these moments are finite).$^\dagger$  Taking $n$ IID draws from this distribution and taking the variance of the sample variance $S_n^2$ gives:
$$\boxed{\mathbb{V}(S_n^2) = \bigg( \kappa - \frac{n-3}{n-1} \bigg) \frac{\sigma^4}{n}}$$
In the special case where the underlying distribution is mesokurtic (e.g., for a normal distribution) we have $\kappa = 3$ and this expression then reduces to:
$$\mathbb{V}(S_n^2) 
= \frac{2n}{n-1} \cdot \frac{\sigma^4}{n} 
= \frac{2 \sigma^4}{n-1}.$$
You might also be interested to note that, in general, the sample variance and sample mean are correlated.  Their covariance is $\mathbb{Cov}(\bar{X}_n, S_n^2) = \gamma \sigma^3/n$ and their corresponding correlation coefficient is:
$$\begin{align}
\mathbb{Corr}(\bar{X}_n, S_n^2) 
&= \frac{\mathbb{Cov}(\bar{X}_n, S_n^2)}{\mathbb{S}(\bar{X}_n) \cdot \mathbb{S}(S_n^2)} \\[6pt]
&= \frac{\gamma \sigma^3}{n} \Bigg/ \frac{\sigma}{\sqrt{n}} \cdot \sqrt{ \Big( \kappa - \frac{n-3}{n-1} \Big) \frac{\sigma^4}{n}} \\[6pt]
&= \frac{\gamma \sigma^3}{n} \Bigg/ \frac{\sigma^3}{n} \cdot \sqrt{\kappa - \frac{n-3}{n-1}} \\[6pt]
&= \frac{\gamma}{\sqrt{\kappa - (n-3)/(n-1)}}, \\[6pt]
\end{align}$$
and as $n \rightarrow \infty$ you get:
$$\mathbb{Corr}(\bar{X}_n, S_n^2) \rightarrow \frac{\gamma}{\sqrt{\kappa - 1}},$$
which is the adjusted skewness of the underlying distribution.  You can find further discussion of moments of the sample moments (including correlation between them) in O'Neill (2014).

$^\dagger$ Actually, you can just assume that the kurtosis is finite, and this implies that all the lower-order moments are also finite.
