One of Maxwell's equations is
$$ \nabla \times \vec{E} = - \frac{d\vec{B}}{dt} \, .$$
Consider an imaginary disk whose normal vector is parallel to the axis of the coil and which is inside the coil.
If you integrate this equation over the area of that disk you get $^{[a]}$
$$\mathcal{E} = - \dot{\Phi}$$
where $\Phi$ is the flux threading the disk and $\mathcal{E}$ is the EMF drop around the loop.
This is called Faraday's law.
So, for each imaginary disk inside your coil we get some EMF as the flux through that disk changes in time.
Now think about the bar magnet's descent.
Suppose we drop it starting way above the entrance to the coil.
It's far away, so there's no flux and no EMF.
As it descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil.
The time changing flux induces some EMF.
This is the initial rise in the red part of the diagram.
As the magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves the time rate of change of total flux increases, so the EMF goes up.
Note that the field lines above and below the bar magnet point in the same direction.
At some point, the bar reaches the middle of the coil.
At this point, the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half.
Therefore, at this point the EMF is zero.
This is the midopint of the diagram where the EMF crosses the horizontal axis.
The falling part of the red section is just the approach to the mid section of the coil.
As the bar magnet exits the coil, more flux is leaving than is entering, so the EMF versus time in the blue section is just the opposite (except for the stretching which you already understand) of the red section.
[a]: On the right hand side, the area integral of the magnetic field is the flux $\Phi$ by definition, and the time derivative just goes along for the ride.
On the left hand side you are doing an area integral of a curl of a vector, which by Stokes's theorem is equivalent to the line integral of the vector itself around the boundary of the area.
The line integral of the electric field vector is the EMF by definition.
TL;DR: The magnet will asymptotically approach a "terminal velocity", at which the magnetic force from the currents in the walls of the tube is exactly balanced by the force of gravity.
To see this, let's denote $F_\text{Lenz}$ as the force from the currents in the wall. This force will obviously depend on the velocity $v$ of the magnet. Its magnitude will increase monotonically with the speed of the magnet (a faster magnet means the flux through a loop of the pipe is changing faster, which results in a larger EMF, which results in more currents in the pipe). This implies that we have
$$
F_\text{tot} = ma = m \frac{dv}{dt} = mg - F_\text{Lenz}(v).
$$
Here, we have defined $v$ to be positive in the downwards direction, while positive values of $F_\text{Lenz}(v)$ correspond to a force in the upwards direction.
To carry this further, we would need to know the precise form of $F_\text{Lenz}(v)$. We can, however, see that the only "stable" solution for a long period of time is $v(t) = v_\text{term}$, where $v_\text{term}$ is the value of the speed which satisfies
$$
F_\text{Lenz}(v_\text{term}) = mg,
$$
i.e., the weight of the magnet is counterbalanced by the force from the currents in the pipe. We know that such a value will exist, since we argued above that $F_\text{Lenz}$ increases with $v$; so for some value of $v$, the Lenz force will be large enough to counter-balance gravity. Moreover, we note that if at any time $v < v_\text{term}$, the velocity will increase (since $mg > F_\text{Lenz}$), driving it towards $v_\text{term}$. Similarly, if at any time $v > v_\text{term}$, the velocity will decrease back towards $v_\text{term}$. Taken all together, this means that after a long period of time, the magnet will approach some terminal velocity.
Finally, note that the magnet cannot stop in the tube: if $v = 0$, then the Lenz force will vanish, and so the only force on the magnet at such a moment will be gravity. It will then be accelerated downwards by gravity and no longer be at rest.
As an aside: an exercise in Zangwill's Modern Electrodynamics (2013) actually presents the calculation of this terminal velocity as a (relatively involved) exercise. Several simplifying assumptions must to be made to solve this problem, most notably that the magnet is a perfect dipole, the walls of the pipe are much thinner than the radius of the pipe, and the self-inductance of the pipe can be ignored. In the end, the result is that
$$
v_\text{term} \propto \frac{a^4 M g}{\mu_0^2 m^2 \sigma d}
$$
where $M$ is the mass of the magnet, $m$ is its dipole moment, $a$ is the radius of the pipe, $d$ is the thickness of the pipe's walls, and $\sigma$ is the conductivity of the pipe. (Finding the precise proportionality factor is left as an exercise for the reader.)
Best Answer
The break in the circuit will stop electrons from flowing, but it does not stop the EMF from being induced. The EMF is simply canceled out by the electrostatic forces which prevent the electrons from "bunching up".
Without the flow of current, there is no generation of a secondary magnetic field, which is the thing which would normally slow the fall of the magnet. Therefore, the magnet remains in free fall. Because power loss is proportional to current, there is no power loss either, and energy conservation works out as expected.
In a copper tube, the electrons are perfectly free to flow in the azimuthal direction, because there would be no "bunching". In essence, each cross-sectional slice of the pipe forms a closed circuit. It is a topologically different system than a broken coil.
For reference - a more accurate (though wordy) transcription of Lenz's law might be
but this only applies to closed loops. If the loop is broken then current cannot flow, and the rest of the rule goes out the window.