Q:

Computing $E[\|X\|^4]$ for normal X

The following seems to hold for $X$ distributed as 0-centered normal with covariance $\Sigma$, any suggestions how to show this analytically? 
$$E[\|X\|^4] = \operatorname{tr}(\Sigma)^2+2\operatorname{tr}(\Sigma^2)$$

A:

Since $\lVert X\rVert^4=(X^TX)^2$, we have for $X\sim N(0,\Sigma)$,
\begin{align}
E\,[\lVert X\rVert^4]=E\,[(X^TX)^2]&=\operatorname{Var}[X^TX]+(E\,[X^TX])^2
\\&=2\operatorname{tr}(\Sigma^2)+(\operatorname{tr}(\Sigma))^2
\end{align}
The expectation of $X^T X$ for any vector $X$ is relatively straightforward, as shown here.  
The variance of $X^TX$ is a bit more work, but for normal $X$ it can be done by generalizing this approach here. 
For a reference regarding derivation of variance of quadratic forms in general, you can look up Linear Regression Analysis by Seber and Lee (second edition, section 1.5).