The other answers give some information as far as how fast hard drives can run during normal use, but your application doesn't involve normal use, so you might be able to crank some higher speeds out of them. If you want to take a rough theoretical approach to this problem, you can look at the stresses in a spinning disk and compare them to the strength of the hard drive platter material (aluminum in your case, but some platters are made from glass and other materials).
Assuming the aluminum part of the platter is bearing most of the load, and it can be approximated as a solid disk (any hole in the middle is very small), and ignoring stresses generated by the fluid flow in your Tesla Turbine (possibly not a safe assumption, but very complicated otherwise) then the stress in the disk can be calculated as follows (from Engineering Toolbox):
$$
\sigma = \frac{\omega^2 r^2 \rho}{3}
$$
where $\sigma$ is the stress (N/m^2), $\omega$ is the angular velocity (rad/s), $r$ is the radius of the disk (m), and $\rho$ is the density of the disk material (kg/m^3).
Rearrange for $\omega$ and you have:
$$
\omega_{max} = \sqrt{\frac{3\sigma_{yield}}{r^2 \rho}}
$$
where $\omega_{max}$ is the maximum theoretical velocity at which the disk would start to break apart and $\sigma_{yield}$ is the yield stress of the disk material.
Example
If you have a 3.5'' (0.0889 m) disk, made of some aluminum alloy with yield stress $\sigma_{yield} = 24.1 \times 10^6$ Pa (I wasn't sure which alloy is used in HD construction, so I picked the lowest yield strength I could find from Aluminum Alloys), and density $\rho = 2700$ kg/m^3, then you would have a maximum rotation speed of:
$$
\omega_{max} = \sqrt{\frac{3(24.1 \times 10^6)}{(0.0889/2)^2 (2700)}} \approx 3680 \text{ rad/s} \approx 35200 \text{ rpm}
$$
Obviously this is very fast, you wouldn't actually want to run your drives that fast due to many factors. Consider that even below the yield stress materials can stretch due to creep and, as other answers have mentioned, higher speeds will cause increased wear in the bearings and other parts. However, this would be a good way to estimate a hard upper limit on your drive speed.
Given the number of assumptions and approximations I made in this analysis, I would feel comfortable with a safety factor of 4 (i.e. $\sigma_{max} = \sigma_{yield}/4$) which would suggest that my example drive could run with low risk of the platters shattering at a speed of approximately 17,600 rpm, albeit with increased wear.
I suggest you look further into your application to figure out what parameters are relevant for you.