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.)
