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
Torque results in angular acceleration, but says nothing about the speed or in which direction an object is actually rotating. If positive torque is applied, that simply means the object's angular velocity is becoming increasingly positive - it can go from a positive value to a larger positive value (speed up), from a negative value to a positive value (slow, stop, and reverse direction), or from a negative value to a smaller negative value (slow down).
When applying positive torque, if the object's angular velocity was already positive, the object spins faster in counterclockwise direction. If the object's angular velocity was zero, the object begins to spin in the counterclockwise direction. If the object's angular velocity was negative to start with, the mangnitude of object's angular velocity decreases as the object's rotation slows. Applying positive torque to an object with negative angular velocity means the object will slow down, stop, and then start spinning the other way.