It is true if $M$ is a diagonal matrix. Let $M=\left(\begin{array}{cc} m_{1} & 0\\ 0 & m_{2}\\ 0 & 0\end{array}\right)$. Assume that $\left|m_{1}\right|\ge\left|m_{2}\right|$, then for any $x:=\left(\begin{array}{c} x_{1}\\ x_{2}\end{array}\right)\in\mathbb{R}^{2}$ with $\left\Vert x\right\Vert _{3}=1$, $\left\Vert Mx\right\Vert _{3}=\left(\left|m_{1}x_{1}\right|^{3}+\left|m_{2}x_{2}\right|^{3}\right)^{1/3}\geq\left(\left|m_{2}x_{1}\right|^{3}+\left|m_{2}x_{3}\right|^{3}\right)^{1/3}=\left|m_{2}\right|$. Take $V$ to be the space spanned by $\left(\begin{array}{c} 0\\ 1\\ 0\end{array}\right)$ and $\left(\begin{array}{c} 0\\ 0\\ 1\end{array}\right)$, then $\left\Vert M^{T}y\right\Vert _{3/2}\leq\left|m_{2}\right|$ for all $y\in V$ with $\left\Vert y\right\Vert _{3/2}=1$. Likewise for the case $\left|m_{1}\right|\le\left|m_{2}\right|$.

But it is still a question for non-diagonal matrices.

It is true if $M$ is a diagonal matrix. Let $M=\left(\begin{array}{cc} m_{1} & 0\\\ 0 & m_{2}\\\ 0 & 0\end{array}\right)$. Assume that $\left|m_{1}\right|\ge\left|m_{2}\right|$, then for any $x:=\left(\begin{array}{c} x_{1}\\\ x_{2}\end{array}\right)\in\mathbb{R}^{2}$ with $\left\Vert x\right\Vert _{3}=1$, $\left\Vert Mx\right\Vert _{3}=\left(\left|m_{1}x_{1}\right|^{3}+\left|m_{2}x_{2}\right|^{3}\right)^{1/3}\geq\left(\left|m_{2}x_{1}\right|^{3}+\left|m_{2}x_{3}\right|^{3}\right)^{1/3}=\left|m_{2}\right|$. Take $V$ to be the space spanned by $\left(\begin{array}{c} 0\\\ 1\\\ 0\end{array}\right)$ and $\left(\begin{array}{c} 0\\\ 0\\\ 1\end{array}\right)$, then $\left\Vert M^{T}y\right\Vert _{3/2}\leq\left|m_{2}\right|$ for all $y\in V$ with $\left\Vert y\right\Vert _{3/2}=1$. Likewise for the case $\left|m_{1}\right|\le\left|m_{2}\right|$.

But it is still a question for non-diagonal matrices.