Where can I find a table of $z$-domain coefficients for Butterworth filters?
You can't find a table of $z$-domain coefficients that is analogous to the tables of Laplace-domain Butterworth transfer functions because it's impossible. For a second-order Butterworth filter, for example, there is no single corresponding $z$-domain transfer function, in contrast to the continuous-time case. 

The reason: discrete-time transfer functions don't cover an infinitely-long frequency axis like a continuous-time system does. When creating a digital approximation to an analog prototype, techniques like the bilinear transform are used to map the frequency response from the $s$-domain transfer function evaluated along the (infinitely long) $j\omega$ axis to the $z$-domain's (finite-length) unit circle. This obviously can't be done exactly, so the various methods for mapping an analog filter to a digital one are approximations.

One factor that plays into the quality of the approximation is the sample rate of the target digital filter. In the paper you linked, you noted that the sampling interval $T$ cancelled out in the calculations of the digital filter coefficients, which is true. However, the sample interval is still required in order to perform the (necessary) frequency prewarping. Recall the prewarping formula:

$$
\omega_p = \frac{2}{T}\tan{\left(\frac{\omega T}{2}\right)}
$$

While the rest of the $T$'s may cancel out after performing the substitution $s = \frac{2}{T}\frac{z-1}{z+1}$, the $T$ in the argument to the tangent function cannot go away. Therefore, the resulting coefficients will always be a function of the sample rate $T$, so your hope of creating a simple table of digital Butterworth filter coefficients invariant of sample rate is stymied. 

A couple other notes regarding this:


The prewarping step is essential to ensure that particular frequencies of interest (i.e. the cutoff frequency for a Butterworth filter) are preserved through the approximation process. Without prewarping, then the stretching of the frequency response that occurs when mapping from an infinitely-long frequency axis to the unit circle would cause the cutoff to appear at a different frequency than desired in the digital filter.
The cookbook linked in niaren's answer also demonstrates the dependence upon the sample rate: you have to trudge through the equations a bit, but you'll see factors in the listed coefficients like $\cos(\omega_0)$ and so on (this is really just a way of slightly hiding the prewarping process). The cookbook defines $\omega_0$ as:
$$
\omega_0 = \frac{2\pi f_0}{F_s}
$$
so when it all comes down to it, you'll have to define your sample rate and prewarp any frequencies of interest in order to design a Butterworth filter approximation. It's not particularly difficult, as the cookbook shows; I'm not sure what the motivation for avoiding those steps is.