Generally, it isn't the case that the (net, total) magnetic flux threading the conducing loop is constant.
Note that the quote uses the word opposes which you have evidently taken to mean nullify. But in fact, if the solution requires an emf, then there must be a non-constant flux threading the loop.
When the conductive material forming the loop has non-zero resistivity, the solution must satisfy
$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt}$$
and
$$\mathscr E = R i$$
where $R$ is the resistance 'round the loop and $i$ is the current.
But
$$\Phi = \Phi_\textrm{ext} + \Phi_i = \Phi_\textrm{ext} + L i$$
where $L$ is the inductance of the conducting loop.
Combining the above yields
$$R i + L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$
For example, consider the case that $\Phi_\textrm{ext} = \Phi_0$ is constant then the solution is
$$i(t) = I_0e^{-\frac{R}{L}t}$$
$$\Phi(t) = LI_0e^{-\frac{R}{L}t} + \Phi_0$$
$$\mathscr E(t) = -\frac{\mathrm d\Phi}{\mathrm dt} = -\left(-\frac{R}{L} \right)LI_0e^{-\frac{R}{L}t} =Ri(t)$$
So, this is all consistent and notice that the flux is not generally constant though the external flux is.
Now, consider the case that $\Phi_\textrm{ext}$ is increasing linearly with time
$$\Phi_\textrm{ext} = \Phi_0 + \phi t$$
then the particular solution is
$$i(t) = -\frac{\phi}{R}$$
$$\Phi(t) = \left(\Phi_0 -\frac{L}{R}\phi\right) + \phi t$$
$$\mathscr E = -\phi = Ri $$
Again, this is consistent and the flux is not generally constant. However, note that this result depends on $R$ being non-zero.
For the $R = 0$ case, we see that
$$L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$
or
$$\Phi_i = - \Phi_\textrm{ext} + \mathrm{constant}$$
$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt} = 0$$
thus we conclude that, for a perfectly conducting loop, the magnetic flux threading the loop is constant.
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
The induced EMF will generate a current which creates a secondary magnetic field which opposes the change in the original magnetic field
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.
Best Answer
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.)