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 $F_\text{friction}>\mu_\text{static}N$.
The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by:
$$D_t = d_t + \omega(t)\times\tag{1}$$
where $D_t$ is the "total" derivative, i.e. the time derivative in the inertial frame, $d_t$ is the time derivative in the frame fixed to the body.
Since there there are no torques on the body, angular momentum is conserved in the inertial frame so that $D_t L=0$. Therefore we get:
$$d_t L = -\omega(t)\times L\tag{2}$$
So, in the frame fixed to the body, the time derivative of $L$ is given by a cross product with $L$, so the time derivative is always at right angles to $L$. Therefore, $|L|$ is constant, but $L$ can certainly change direction by (2): its head is constrained to a sphere.
So this reasoning yields $I_{1}^2w_{1}^2 + I_{2}^2w_{2}^2 + I_{3}^2w_{3}^2 = const.$
The other equation $I_{1}w_{1}^2 + I_{2}w_{2}^2 + I_{3}w_{3}^2 = 2T_{rot}$ does not follow from the paragraph you referenced, but it is correct nonetheless: it is instead a statement of conservation of rotational kinetic energy, not of angular momentum.
BTW (1) is derived from the Leibnitz rule applied to the matrix equation $X=\exp(H(t))\,Y(t)$ where $H$ is a $3\times 3$ skew-symmetric matrix; $Y$ are the co-ordinates in the frame fixed to the body, $X$ are those in the inertial frame and $\exp(H)$ is the total rotation operator: we think of the co-ordinates written as $3\times 1$ column vectors. At $t=0$ (when our co-ordinate frames are instantaneously aligned) this yields $\dot{X} = \Omega(t)\,Y + \dot{Y}$, where $\Omega(t)$ is derived by a complicated Lie-theoretical formula from the skew symmetric $H$, but $\Omega$ is nonetheless a skew-symmetrix $3\times 3$ matrix, and can thus be represented as a cross product $\omega\times Y$.
Best Answer
For the first question, you're talking about a constraint, an axle in this case. If the axle isn't parallel to a principal axis, it imposes oscillating torques on the body about the axes perpendicular to the axle. The angular momentum of the body thus oscillates.
For the second, you get torque-free precession. The angular momentum remains constant, but the rotation vector doesn't. So, the rotating body wobbles. In the absence of friction and external torques, this persists indefinitely. However, if the body has internal moving parts subject to friction they may damp the wobble. The result is a body rotating around its axis of maximum inertia. This is the principle behind "nutation dampers" on spacecraft. This effect was a surprise to early spacecraft engineers.