Variance of the $\hat{\sigma}$ of a Maximum likelihood estimator With a random sample having size $n$ from a normal distribution $\mathcal{N}(\mu,\sigma^2)$, and the ML estimate of $$g(\mu,\sigma^2) = \mu +\sigma$$ being $$\widehat{g(\mu,\sigma^2)}=\overline{x}+\hat{\sigma} \:\:\:\:\:\bigg(\text{where}\:\:\hat{\sigma}=\sqrt{\hat{\sigma}^2}\;\bigg)$$ I need to find an estimate of the standard deviation of the estimate of $g(\mu,\sigma^2)$.
In order to do this, I think I need to "simply" find Var$(\overline{x}) +$ Var$(\hat{\sigma})$
Var$(\overline{x})$ I am not too concerned about, but I am a little stumped on how to find Var$(\hat{\sigma})$. From the help here, I think to use the fact that
$$\frac{n\hat{\sigma}^2}{\sigma^2}\sim \chi_{n-1}^2$$
Following this through, I get down to $$Var(\hat{\sigma}^2)=\frac{2\sigma^4(n-1)}{n^2}$$
From here I am a little bit stuck though - can I simply just take the square root of both sides, take the square root inside the variance, and then have my result? I was kind of hoping to not see $\sigma$ appearing in the function for variance, but I suppose it has to?

You will not get the correct variance of the standard deviation by working out the variance of the variance and taking its square root $-$ that's the population standard deviation of the sample variance not the population variance of the sample standard deviation. You also have to be careful about what $\hat{\sigma}$ is!
At the normal, the MLE of $\sigma$ is $s_n$, the $n$-denominator version of sample standard deviation (rather than the Bessel-corrected $s_{n-1}$).
$$E(s_n) = E\left(\sqrt{\frac{n-1}{n}}s_{n-1}\right) = \sqrt{\frac{n-1}{n}}E(s_{n-1})$$
and $E(s_{n-1})$ can be found in Macro's answer here
So now
\begin{eqnarray}
\text{Var}(s_n) &=& E(s_n^{2}) - E(s_n)^2 \end{eqnarray}
and the rest is substitution (take care with the first term as well as with the second).
The variance result will have $\sigma^2$ multiplied by a constant (being a function of $n$ only).
