How does one calculate or even approximate the force observed by a magnet as its dropped inside of a diamagnetic cylinder? How does one calculate or even approximate the force observed by a magnet as its dropped inside of a diamagnetic cylinder? 
Wikipedia and Google does not provide much information on this topic. Wikipedia suggest to calculate the change in magnetic energy with respect to position, however I am unsure how the velocity should play a role as well.
 A: In diamagnetic material a magnetic dipole moment will develop in presence of external field. Magnetic dipole moment per volume is:
$$\mathbf{M}=\chi\mathbf{H}= {\chi\mathbf{B} \over \mu}={1 \over \mu_0}{\chi \over \chi+1}\mathbf{B}$$
In a non-uniform magnetic field there would be a force on a magnetic dipole:
$$\mathbf{F}=\nabla (\mathbf{m}\cdot\mathbf{B})$$
One could integrate this over the cylinder to find the force:
$$\mathbf{dF}=\nabla(\mathbf{dm}\cdot\mathbf{B})=\nabla(\mathbf{M}\cdot\mathbf{B})dv={1 \over \mu_0}{\chi \over \chi+1}\nabla(B^2)dv$$
$$\mathbf{F}={1 \over \mu_0}{\chi \over \chi+1}\int_{V_{cylinder}} \nabla(B^2)dv$$
But we still need magnetic field of the magnet. We can assume a uniform magnetic moment per volume $\mathbf{M}$ for it and find the field by integrating dipole field over the magnet volume:
$$\mathbf{B(\mathbf{r})}={\mu_0 \over 4\pi} \int_{V_{magnet}} [{3 (\mathbf{r}-\mathbf{r'})(\mathbf{M}\cdot(\mathbf{r}-\mathbf{r'}))\over (\mathbf{r}-\mathbf{r'})^5}-{\mathbf{M}\over (\mathbf{r}-\mathbf{r'})^3}] dv$$
This does not include velocity dependent force due to eddy currents mentioned in the comments.
Update I'll post the solution of a similar (and simplified) problem which you could later extend to your own problem.
Here we have a ring instead of cylinder with inductance L, resistance R, radius r, and mass m. And instead of magnet dropping we assume that the ring is dropping toward the magnet. 
To make it simpler again we assume a linear z dependence for $B_z$ component of the magnet's magnetic field: $B_z = B_0(1-\alpha z)$
Due to symmetry $B_\phi=0$ so $\nabla\cdot\mathbf{B}=0$ implies that ${\partial{B_\rho}\over{\partial \rho}}=B_0 \alpha$ then $B_\rho(\rho)=B_0 \alpha \rho + C$
Since $B_\rho(\rho=0)=0$ we have $B_\rho(\rho)=B_0 \alpha \rho$ and $B_\rho(r)=B_0 \alpha r$
Taking up as positive z direction and counter-clockwise as positive direction for current in the ring:
$$ \tag{1}
F = m \ddot z = -2\pi r I B_\rho(r) - m g = -2\pi r^2 \alpha I B_0 -mg
$$
As for current we have:
$$
\phi = \pi r^2 B_0 (1-\alpha z)\\
\epsilon_{emf} = -\dot \phi = \pi r^2 B_0 \alpha \dot z\\
\epsilon_{emf} - R I - L \dot I = 0$$ $$
\pi r^2 B_0 \alpha \dot z - R I - L \dot I = 0 \tag{2}
$$
Now we find $I$ and $\dot I$ from Eq. 1 and substitute in Eq. 2
$$
\pi r^2 B_0 \alpha \dot z - R {m(\ddot z + g) \over 2\pi r^2 \alpha B_0} - L {m\dddot z \over 2\pi r^2 \alpha B_0} = 0\\
\dddot z + {R \over L} \ddot z + {2 \pi^2 r^4 \alpha^2 B_0^2 \over m L} \dot z + {R \over L}g = 0\\
v \equiv \dot z \quad
\omega_0^2 \equiv {2 \pi^2 r^4 \alpha^2 B_0^2 \over m L} \quad
\gamma \equiv {R \over 2L} \quad {\omega_0^2 v_t} \equiv {R \over L}g\\
\ddot v + 2\gamma \dot v + \omega_0^2 v + \omega_0^2 v_t = 0
$$
Now you can see that the velocity has a damped oscillation around the terminal velocity.
