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
Take the right pedal as an example. It uses a right-hand thread, so turning the pedal spindle clockwise (CW) relative to the crank will screw the spindle in, counter-clockwise (CCW) will unscrew it.
Say you put the bike in a repair stand, grab the right pedal and gently simulate the motion of someone riding the bike (always keeping the pedal platform horizontal as if your foot were there). The body of the pedal will turn CCW relative to the crank.
Intuitively you would think this might help unscrew the pedal! But in reality the clamping torque of the threads will be far, far greater than any friction in the bearings could generate. Even if the bearings were to seize, it would be very difficult to unscrew the pedal (unless it was never tight to begin with).
Now consider someone riding the bike, putting weight on the right pedal. This applies a force perpendicular to the ground, no matter where the foot is in the pedal stroke. Relative to the crank arm, this radial force rotates CCW, which - via the process of mechanical precession - creates a CW torque on the pedal spindle (thus tightening it). The rotations on the left side are all reversed, so it must use the opposite threading to prevent the pedal from coming loose.
Also see "Precession" on Wikipedia, from where the below illustration was taken (CC-SA).