When $\sigma=1$, this is the first moment of Chi-distribution. Furthermore, \begin{equation}
\mathbb{E}_{x\sim\mathcal{N}(0,\sigma^2I)}\{ \|x\|_2^k \} = \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \|\sigma x\|_2^k \}= \sigma^k \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \| x\|_2^k \},
\end{equation}
where $\mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \| x\|_2^k \}$ is the $k$th moment of Chi-distribution, which has value
\begin{equation}
\mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \|x\|_2^k \} = 2^{k/2} \frac{\Gamma((N+k)/2)}{\Gamma(N/2)}.
\end{equation}

Two comments:
1. I think $x\sim\mathcal{N}(0,\sigma I_N)$ should be replaced by $x\sim\mathcal{N}(0,\sigma^2I_N)$ in the original post.
2. This result is different from that of Botev's.
Let me know if I made any mistakes.