Following suggestions on Stack Exchange, we use the homothety with respect to $O$ and factor $-3$. If $X$ is a point, then $X'$ denotes the point for which $\overrightarrow{XO}= -3 \overrightarrow{X'O}$. So $M=N'$ and $A'$ is the midpoint of the face opposite to $A$. One has $XY=3X'Y'$ and the desired inequality becomes $3MA'+3MB'+3MC'+3MD'\geq MA+MB+MC+MD$.
It thus suffices to prove inequalities of the type $MB'+MC'+MD'\geq MA$. Thus put $f(X)=XB'+XC'+XD'-XA$. We have to show that $f(M)\geq0$.
Let us fix our tetrahedron as follows.
$A=\left(0,0,\sqrt{\frac{2}{3}}-\frac{1}{2 \sqrt{6}}\right)$,
$B= \left(-\frac{1}{2 \sqrt{3}},-\frac{1}{2},-\frac{1}{2 \sqrt{6}}\right)$,
$C=\left(-\frac{1}{2 \sqrt{3}},\frac{1}{2},-\frac{1}{2 \sqrt{6}}\right)$,
$D=\left(-\frac{1}{2 \sqrt{3}},\frac{1}{2},-\frac{1}{2 \sqrt{6}}\right)$.
Put $f_1(X)=XB'+XC'$, $f_2(X)=XD'-XA$, so that $f(X)=f_1(X)+f_2(X)$.
Notice that $f(O)=0$ and notice that $f(M)$ is positive if $M$ is far away from $O$.
Now take $M$ so that $f(M)$ is an absolute minimum. Note that $f$ is positive at $A$, $B'$, $C'$, $D'$, so that $M$ is none of those points. We have $f_2(M)<0$ and the gradients of $f_1$, $f_2$ cancel each other at $M$. That means that the level sets through $M$ of $f_1$ and $f_2$ touch at $M$. (There is no point where both gradients vanish.)
The level set through $M$ of $f_1$ is an ellipsoid, with $f_1$ smaller inside, and the level set through $M$ of $f_2$ is the lower sheet of a hyperboloid of two sheets, with $f_2$ more negative inside the sheet. As each level set is the curved boundary of a convex region, there is no other point where the two level sets touch.
The level sets are symmetric with respect to the plane $ \left\{X\mid XB'=XC' \right\}$, so $M$ must lie on that plane. We have shown that $MB'=MC'$. Similarly $MC'=MD'$ and $M$ must be of the form $(0,0,x)$, so that $f(M)=\frac{\sqrt{72 x^2-4 \sqrt{6} x+3}-\sqrt{8 x^2-4 \sqrt{6} x+3}}{2 \sqrt{2}}\geq0$.