Let us interpret the phrase "whose entries are taken from a gaussian distribution" as implying that the entries of the random matrix $M$ are iid. Then the key is the simple observation that, for any orthogonal matrix $Q$, the matrix $QM$ equals $M$ in distribution. Let now $Q$ be any orthogonal matrix such that $xQ=e_1:=[1,0,\dots,0]$. Then $xM$ equals \begin{equation} xQM=e_1M=[M_{1,1},\dots,M_{1,n}] \end{equation} in distribution. So, $\|xM\|$ equals $\sqrt{\frac2n\,X}$ in distribution, where $X\sim\chi^2_n=\text{Gamma}(\frac n2,2)$. It follows that \begin{equation} E\|xM\|=\frac{2\Gamma(\frac{n+1}2)}{\sqrt n\,\Gamma(\frac n2)}, \end{equation} which is asymptotic to $\sqrt 2=\sqrt{E\|xM\|^2}$ for large $n$, as should be expected, in view of the measure concentration phenomenon.

The key is the simple observation that, for any orthogonal matrix $Q$, the matrix $QM$ equals $M$ in distribution. Let now $Q$ be any orthogonal matrix such that $xQ=e_1:=[1,0,\dots,0]$. Then $xM$ equals \begin{equation} xQM=e_1M=[M_{1,1},\dots,M_{1,n}] \end{equation} in distribution. So, $\|xM\|$ equals $\sqrt{\frac2n\,X}$ in distribution, where $X\sim\chi^2_n=\text{Gamma}(\frac n2,2)$. It follows that \begin{equation} E\|xM\|=\frac{2\Gamma(\frac{n+1}2)}{\sqrt n\,\Gamma(\frac n2)}, \end{equation} which is close to $\sqrt 2=\sqrt{E\|xM\|^2}$ for large $n$, as should be expected, in view of the measure concentration phenomenon.

The key is the simple observation that, for any orthogonal matrix $Q$, the matrix $QM$ equal $M$ in distribution. Let now $Q$ be any orthogonal matrix such that $xQ=e_1:=[1,0,\dots,0]$. Then $xM$ equals \begin{equation} xQM=e_1M=[M_{1,1},\dots,M_{1,n}] \end{equation} in distribution. So, $\|xM\|$ equals $\sqrt{\frac2n\,X}$ in distribution, where $X\sim\chi^2_n=\text{Gamma}(\frac n2,2)$. It follows that \begin{equation} E\|xM\|=\frac{2\Gamma(\frac{n+1}2)}{\sqrt n\,\Gamma(\frac n2)}. \end{equation}

Let us interpret the phrase "whose entries are taken from a gaussian distribution" as implying that the entries of the random matrix $M$ are iid. Then the key is the simple observation that, for any orthogonal matrix $Q$, the matrix $QM$ equals $M$ in distribution. Let now $Q$ be any orthogonal matrix such that $xQ=e_1:=[1,0,\dots,0]$. Then $xM$ equals \begin{equation} xQM=e_1M=[M_{1,1},\dots,M_{1,n}] \end{equation} in distribution. So, $\|xM\|$ equals $\sqrt{\frac2n\,X}$ in distribution, where $X\sim\chi^2_n=\text{Gamma}(\frac n2,2)$. It follows that \begin{equation} E\|xM\|=\frac{2\Gamma(\frac{n+1}2)}{\sqrt n\,\Gamma(\frac n2)}, \end{equation} which is asymptotic to $\sqrt 2=\sqrt{E\|xM\|^2}$ for large $n$, as should be expected, in view of the measure concentration phenomenon.