First lets give the box some dimensions: $w$ for width and $h$ for height.
A level, stationary box
The scenario you described above is almost correct if $F\,h\leq m\,g\,w$.
The net torque on the box would be $F\frac{h}2 - N\frac{w}2$, which if $N=m\,g$ would result in a net counter clockwise (negative in my chosen reference frame) torque. This would mean the corner not in red would be pushed into the floor, supporting some of the weight of the box, reducing $N$ until $F\frac{h}2 - N\frac{w}2 = 0$.
An accelerating box
If there is a net torque on the box (i.e. $F\frac{h}2 - N\frac{w}2 \geq 0$) then the box will be accelerating around the red corner. Note that to accelerate around the red corner would accelerate the center of mass. Since it's accelerating we can no longer claim that the net forces are zero. In particular now $F\gt F_\text{friction}$ and $N\gt m\,g$.
So Your first and second questions are answered by the fact that yes $N$ will increase as the box starts rotating, but that increase will allow it to overpower the weight of the box allowing the center of mass to accelerate upwards. Once the center of mass has moved upwards, there is now room for the corner to rotate without penetrating the ground.
As for your third question. No $F\neq F_\text{friction}$ once the box starts accelerating. $F_\text{friction}$ will reduce once the box starts rotating.
For your forth question, the net linear acceleration of the CG is not zero. You are correct that the CG describes an arc of a circle.
If you would like to calculate these values I would proceed as follows:
The moment of inertia of a box about its corner is
$$I=m\frac{w^2+h^2}3 \, .$$
The angular equivalent of $F=ma$ is
$$\tau=I\,\alpha=I\,\dot\omega=I\,\ddot\theta \, .$$
Now before we looked at the net torque about the center of gravity. While it's possible to solve this problem using that origin, choosing the red dot as our center allows us to skip a few steps:
\begin{align}
\tau &= F\frac{h\,\cos(\theta)+w\,\sin(\theta)}2-m\,g\frac{w\,\cos(\theta)-h\,\sin(\theta)}2 \\
\ddot\theta &= F\frac{h\,\cos(\theta)+w\,\sin(\theta)}{2I}-m\,g\frac{w\,\cos(\theta)-h\,\sin(\theta)}{2I} \, .
\end{align}
Unfortunately, this is equivalent to the large oscillation pendulum problem by a rotation of coordinate system. As such there is no analytic solution, but a numerical solution could get you $\theta(t)$.
Of course this solution would only be valid up to the point where friction would give way $F_\text{friction}>\mu_\text{static}N$.
Best Answer
OMG the solution was as simple as imagining the corner as the center of the "torque circle" therefore the radius to mg and fn is indead l and is 2R-h to the applied force.