Multivariate normal concentration If $X\sim N(0,\Sigma)$ for some $d$-dimensional normal distribution, then $X = \Sigma^{1/2} Z$ where $Z\sim (0,I)$. How to compute the following quantity?
$$
\operatorname{var}  (X^T X)
=
\operatorname{var}(Z^T \Sigma Z)
$$
Moreover, if we are in the growing dimension regime, under what condition posed on $\Sigma$ (such as restricted strong convexity) do we have as $d\to\infty$ the following?
$$
\operatorname{var}\left( X^\top X \right)
=
o(d^2)
$$
Update: do we have some sort of concentration inequality for $X^T X$?
 A: Let $ Y := Z^T \Sigma Z $. We have $ \operatorname{var}(Y) = \mathbb{E}(Y^2) - \mathbb{E}(Y)^2 $ and $ \mathbb{E}(Y) = \sum_i \sigma_{i, i} $. We thus need to compute $ \mathbb{E}(Y^2) $. For this, write
$$  \mathbb{E}(Y^2) = \mathbb{E}\left( \sum_{i, j, k, \ell} Z_i Z_j Z_k Z_\ell \sigma_{i, j} \sigma_{k, \ell}  \right) =  \sum_{i, j, k, \ell} \mathbb{E}\left(Z_i Z_j Z_k Z_\ell \right)  \sigma_{i, j} \sigma_{k, \ell}   $$
Due to $ \mathbb{E}\left(Z_i \right) = 0  $, the only non zero contribution of this last expectation comes from the indices such that $ i = j \neq k = \ell $ (plus the permutations of this case) or $ i = j = k = \ell $. In the first case, we have $ \mathbb{E}\left(Z_i^2 \right) = 1 $ and in the second case, $ \mathbb{E}\left(Z_i^4 \right) = 3$. 
In the case $ \pi_1 := (i = j \neq k = l ) $ the sum becomes $ (\sum_{i \neq k} \sigma_{i, i} \sigma_{k, k} )^2 = (\sum_i \sigma_{i, i})^2 - \sum_i \sigma_{i, i}^2 $ and in the cases $ \pi_2 := (i = k \neq j = l) $ and $ \pi_3 := (i = l \neq j = k) $, it is equal to $ \sum_{i \neq j} \sigma_{i, j}^2 = \sum_{i,j} \sigma_{i, j}^2 - \sum_i \sigma_{i, i}^2 $. Summing, we get 
$$ \mathbb{E}(Y^2) = 2 \sum_{i, j} \sigma_{i, j}^2 + \left( \sum_i \sigma_{i, i} \right)^2 - 3 \sum_i \sigma_{i, i}^2 + 3 \sum_i \sigma_{i, i}^2  $$
hence, the variance is
$$ \operatorname{var}(Y) = 2 \sum_{i, j} \sigma_{i, j}^2  $$
Concerning your question on the regime $ d \to \infty $, you can for instance suppose that $ \max_{i } \sigma_{i, j}^2 = o(1) $ to get $\operatorname{var}(Y) = o(d) $ (with the trivial inequality $ \sum_i \sigma_{i, j}^2 \leq d \max_i \sigma_{i, j}^2 $). You can also put hypotheses directly on the quantity of interest, namely $ \sum_{i, j} \sigma_{i, j}^2 $.
A: For the variance, you need to evaluate $E[X^T X]^2$ and $E[(X^T X)^2]$.
For the former:
$$E[X^T X] = \operatorname{tr} E[X^T X] = \operatorname{tr} E[X X^T] = \operatorname{tr} \Sigma$$
For the latter, using Isserlis' theorem:
$$
\begin{aligned}
&E[(X^T X)^2] = \sum_{ij} E[X_i^2 X_j^2] \\
&= \sum_{ij} \bigl\{E[X_i X_i] \, E[X_j X_j] + 2 \, E[X_i X_j] \, E[X_i X_j]\bigr\} \\
&= \sum_{ij} \bigl\{\Sigma_{ii} \, \Sigma_{jj} + 2 \, \Sigma_{ij}^2\bigr\} \\
&= (\operatorname{tr} \Sigma)^2 + 2 \operatorname{tr} \Sigma^2
\end{aligned}
$$
Hence $\operatorname{Var}[X^T X] = 2 \operatorname{tr}\Sigma^2 = 2 \sum_{i=1}^{d} \lambda_i^2$, i.e. twice the sum of the squared eigenvalues.
