One way to get the answer is to use the moment generating function of a Gaussian distribution.
Note that:
\begin{equation}
\begin{split}
\int_\mathbb{R}^n \exp\{-\frac{x^Tx}{2}\}||x||^{2N}_2 & = (2\pi)^{\frac{n}{2}}\int_\mathbb{R}^n \frac{1}{(2\pi)^{\frac{n}{2}}}\exp\{-\frac{x^Tx}{2}\}||x||^{2N}_2\\
& = (2\pi)^{\frac{n}{2}}\int_\mathbb{R}^n \frac{1}{(2\pi)^{\frac{n}{2}}}\exp\{-\frac{x^Tx}{2}\}(x_1^2 + ... x_n^2)^N\\
& = (2\pi)^{\frac{n}{2}}E[(x_1^2 + ... x_n^2)^N]\\
\end{split}
\end{equation}
By the multinomial theorem $(x_1^2+...+x_n^2)^N = \sum_{|\alpha| = N}\binom{n}{\alpha} \prod_{i=1}^n x_i^{\alpha_i}$.
Placing this expression in the previous expectation and swapping the expectation and finite sum gives us:
\begin{equation}
\begin{split}
(2\pi)^{\frac{n}{2}}E[(x_1^2 + ... x_n^2)^N] &= (2\pi)^{\frac{n}{2}}E[\sum_{|\alpha| = N}\binom{n}{\alpha}\prod_{i=1}^n x_i^{2\alpha_i}]\\
&= (2\pi)^{\frac{n}{2}}\sum_{|\alpha| = n}\binom{n}{\alpha}E[\prod_{i=1}^n x_i^{2\alpha_i}]
\end{split}
\end{equation}
Note that $E[\prod_{i=1}^n x_i^{\alpha_i}]$ is the $\alpha$ moment of $X$, and can be obtained by differentiating the moment generating function of a mean 0 covariance $I$ multivariate Gaussian. $M_X(t) = \exp\{\frac{1}{2}t^Tt\}$, thus $E[\prod_{i=1}^n x_i^{\alpha_i}] = \frac{\partial}{\partial^\alpha x}\exp\{\frac{1}{2}t^Tt\}|_{t=0}$.
This derivative evaluated at $0$ is given by:
\begin{equation}
\frac{\partial}{\partial^\alpha x}\exp\{\frac{1}{2}t^Tt\}|_{t=0} = \prod_{i=1}^n \prod_{j=1}^{\alpha_i}(2j-1)
\end{equation}
Putting this expression in the previous expression gives us:
\begin{equation}
\begin{split}
(2\pi)^{\frac{n}{2}}\sum_{|\alpha| = N}\frac{N!}{\alpha_1!...\alpha_n!}\prod_{i=1}^n \prod_{j=1}^{\alpha_i}(2j-1)
\end{split}
\end{equation}
Using the fact thatthere is a closed form for the first k odd numbers, we have that the far right product is equal to $\frac{(2\alpha_i)!}{2^{\alpha_i}\alpha_i!}$, giving us:
\begin{equation}
\begin{split}
(2\pi)^{\frac{n}{2}}\sum_{|\alpha| = N}\frac{N!}{\alpha_1!...\alpha_n!}\prod_{i=1}^n \prod_{j=1}^{\alpha_i}(2j-1) & = (2\pi)^{\frac{n}{2}}\sum_{|\alpha| = N}\frac{N!}{\alpha_1!...\alpha_n!}\prod_{i=1}^n \frac{(2\alpha_i)!}{2^{\alpha_i}\alpha_i!}\\
& = (2\pi)^{\frac{n}{2}}\sum_{|\alpha| = N}\frac{N!\prod_{i=1}^n\binom{2\alpha_i}{\alpha_i}}{2^N}\\
& = (2\pi)^{\frac{n}{2}}\frac{N!}{2^N}\sum_{|\alpha| = N}\prod_{i=1}^n\binom{2\alpha_i}{\alpha_i}\\
& = (2\pi)^{\frac{n}{2}}\frac{N!}{2^N}\binom{2N}{N}\\
\end{split}
\end{equation}
With the last equality following from equation 1 in this research article giving us that $\sum_{|\alpha| = N}\prod_{i=1}^n\binom{2\alpha_i}{\alpha_i} = \binom{2N}{N}$.
This isn't obviously equivalent to what you wrote down, but seems to be consistent with the coefficients of the first 4 noncentral moments provided here for $n = 1$.
EDIT:
I think there may be a mistake in me referencing the research article. If you refer to Alan Yan's comment on this answer it seems to give the appropriate value. That doesn't explain why the coefficients are the coefficients of the non-central moments, but if I figure out what's going on I can make an edit to the answer.
