[Edit:]

I think we maybe able to treat this using the duality of the $\ell_3$ and $\ell_{3/2}$ norms.

Observe that it suffices to prove

$$ \inf_{V^2\subset\mathbb{R}^3} \sup_{y\in V, \|y\|_{3/2} = 1} \|M^Ty\|_{3/2} \leq \inf_{\|x\|_3 = 1, x\in \mathbb{R}^2} \|Mx\|_3 $$

For a fixed $y$,
$$\|M^Ty\|_{3/2} = \sup_{\|x\|_3 = 1} y^TMx$$
For fixed two-space $V$, let $P_V$ denote the orthogonal (rel. $\ell_2$ on $\mathbb{R}^3$) projection to $V$. Then
$$ \sup_{y\in V, \|y\|_{3/2} = 1} \|M^Ty\|_{3/2} = \sup_{\|y\|_{3/2} = 1} \sup_{\|x\|_3 = 1} y^TP_VMx$$
We can swap the two supremums as the two unit spheres are both compact. So we have
$$ \sup_{y\in V, \|y\|_{3/2} = 1} \|M^Ty\|_{3/2} = \sup_{\|x\|_3 = 1} \|P_VMx\|_3$$

Now take $V$ to be the two-space spanned by $v,n$, where $n$ is in the kernel of $M^T$ (which we assume to be only 1 dimensional; if the dimension is bigger, then setting $V$ to be this kernel your inequality is trivial), and $v = Mx_0$ achieves the infimum
$$\inf_{\|x\|_3 = 1} \|Mx\|_3$$

Now, the unit ball $B\subset\mathbb{R}^2$ in $\ell_3$ is convex. So is its image under $M$. So the question reduces to whether the supporting hyperplane to $MB$ at $v$ is $\ell_2$ orthogonal to $v$. In the diagonal case this is trivially true. In general I am a bit stuck at the moment.

**below is completely wrong. Kept to make the comments to this answer make sense**

Let $M = \begin{pmatrix} k&0 \\ k&0 \\ 0 &0\end{pmatrix}$. Then $\inf_{\|x\|_3 = 1}\|Mx\|_3 = 0$. Whereas for any two dimensional subspace $V$, $0 < \sup_{\|y\|_{3/2} = 1, y\in V}\|M^T y\|_{3/2} \propto k$. So the proposed statement is false.