What are the values of the derivative of Riemann's zeta function at the known non-trivial zeros?

Here is an answer in a few parts.

1. Ghaith Hiary has computed fairly large tables of zeros at large height, which are at

   http://sage.math.washington.edu/home/hiaryg/page/index.html

   I believe that each table has 10 million zeros starting at t = 10^n, for n from 12 to 28. There are tables of derivatives of Z'(t) at the zeros, but not zeta'(1/2 + it). At at zero, it is not hard to go from one to the other, though.

2. To compute what you want to compute, I think the formula you wrote down is not going to be computationally practical due to the exponential decay of the gamma function and the exponential growth of the sine function. (I think may already be precision issues in the picture in Stopple's answer.) If I calculate correctly (I hope I'm not embarrassing myself), then $$ -\frac{\zeta'(1/2 + i\gamma)}{\zeta'(1/2 - i\gamma)} = e^{-2 i \theta(\gamma)} $$ where $\theta$ is the Riemann-Siegel theta function. This is probably a better way to compute the quotient.

   To make the above picture using sage, if you have the optional package database_odlyzko_zeta installed, you can use:

   
   import mpmath
   mpmath.mp.prec = 200
   
   L = []
   for gamma in zeta_zeros()[:100]:
       z = exp(-2 * CC.0 * CC(mpmath.siegeltheta(gamma)))
       L.append((z.real(), z.imag()))
   
   P = points(L)
   P.save('plot.png', figsize=[5,5])
   
   

   Something similar will work if you want to parse Hiary's files.

3. As for whether there is any tendency for these numbers to be close to -1: I think there is not. This looks like the question of whether there is any asymptotic relation between the zeros of the zeta function and gram points, and I think that the answer is expected to be no, but I don't know if that is a theorem. (Those thoughts should not stop you from checking, though.)

To elaborate a little more, here's some Mathematica code:

ListPlot[Table[{rho = ZetaZero[n];z = N[2 (2 Pi)^(rho - 1) Gamma[1 - rho] Sin[Pi rho/2]]; {Re[z], 
Im[z]}}, {n, 1, 100}], AspectRatio -> Automatic]

Here's the output:

     (source)