Consider the random vector $X=(X_1,\dots,X_k)$ such that $X\sim N_k(\mu,\Sigma)$.
Set $Q(X)=X^TAX=\sum_{i=1}^kX_i^2$, where $A=\mathrm I_k$. A good reference for the study of $Q(X)$ is "Quadratic forms in random variables" by Mathai and Provost.
Specifically, denote by $P$ an orthogonal matrix that diagonalizes $\Sigma$, and write $P^T\Sigma P=\mathrm{Diag}(\lambda_1,\dots,\lambda_k)$. Also define
$$
b=P^T\Sigma^{-1/2}\mu.
$$
Then,
$$
Q(X)=\sum_{i=1}^k\lambda_i\left(U_i+b_i\right)^2,\qquad(1)
$$
where $U_i\sim N_k(0,\mathrm I_k)$ (equation $(4.1.1)$ of the reference). This is sometimes called a generalized Chi-squared distribution, and the length of the vector $\sqrt{Q(X)}$ is thus called generalized Chi distribution. The Laplace transform of $Q(X)$ is also obtained in equation $(4.2b.6)$ as
$$
\mathrm L(s)=\exp\left(-\frac12\sum_{i=1}^kb_i^2\right)\exp\left(\frac12\sum_{i=1}^kb_i\frac1{1+2s\lambda_i}\right)\prod_{i=1}^k\frac1{\sqrt{1+2s\lambda_i}},
$$
for $\left|2s\lambda_i\right|<1$. Now, in general $(1)$ does not follow a well-known distribution. As you mention in OP, if $\Sigma=\sigma^2\mathrm I_k$ and $\mu\neq0$, then $Q(X)$ follows a non-central chi-squared distribution. Another case that has a closed-form solution is if $\mu=0$, and $\Sigma$ is diagonal with elements $\sigma_1^2,\dots,\sigma_k^2$. Then (c.f. wikipedia as well as $[5]$ therein), if $\sigma_i\neq\sigma_j$, the density of $Q(X)$ can be computed as
$$
f(x)=\sum_{i=1}^{k} \frac{e^{-\frac{x}{\sigma_i^2}}}{\sigma_i^2 \prod_{j=1, j\neq
i}^{k} (1- \frac{\sigma_j^2}{\sigma_i^2})}1_{x\ge0}.
$$
From this you can deduce the distribution of $\sqrt{Q(X)}$. Similar calculations can be done if $\sigma_i=\sigma_j$ for some $i,j\in\{1,\dots,k\}$.
To answer your question is a general setting is more difficult since to my knowledge, these distributions do not follow other known laws. Therefore, it depends on what application you have in mind. If you want to calculate moments, then you might be able to exploit formula $(1)$. If you want to approximate the pdf of $\sqrt{Q(X)}$, then you can use some of the asymptotic expansions of Chapter $4$ of the book that I mentioned in the beginning of this post.
We can avoid using the fact that $X^2\sim\sigma^2\chi_1^2$, where $\chi_1^2$ is the chi-squared distribution with $1$ degree of freedom, and calculate the expected value and the variance just using the definition. We have that
$$
\operatorname E X^2=\operatorname{Var}X=\sigma^2
$$
since $\operatorname EX=0$ (see here).
Also,
$$
\operatorname{Var}X^2=\operatorname EX^4-(\operatorname EX^2)^2.
$$
The fourth moment $\operatorname EX^4$ is equal to $3\sigma^4$ (see here). Hence,
$$
\operatorname{Var}X^2=3\sigma^4-\sigma^4=2\sigma^4.
$$
Best Answer
Note that $X_1=\frac{X-\mu}{\sigma^2}$ follows the standard normal distribution, and so $X_1^2$ has the $\chi^2$ distribution.
Now, $X_1^2=(X^2-2\mu X+\mu^2)/\sigma^4$. Thus you can get the distribution of $X^2$ in terms of distributions of $X_1^2$, $X$ and a constant.
I do not know though, if the distribution of $X^2$ has any standard name or not.