The result in the OP

$$\sum_{n=1}^\infty \sum_{k=1}^\infty\frac{G_n}{k}\frac{\Gamma(k+\frac{3}{2})\Gamma(n+\frac{1}{2})}{(n+k+1)!}=\frac{\pi(1-8\log 2)}{8}=-1.78489,$$
does not seem to agree with a numerical estimate of the series. I first evaluated the sum over $k$, which has a closed form expression, 
$$\sigma_n=\sum _{k=1}^{\infty } \frac{\Gamma \left(k+\frac{3}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{k (k+n+1)!}=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2}\right) \left(\psi ^{(0)}(n+2)-\psi ^{(0)}\left(n+\frac{1}{2}\right)\right)}{2 \Gamma (n+2)},$$
with $\psi^{(0)}$ the polygamma function, and then evaluated
$$S_N=\sum_{n=1}^N G_n \sigma_n,$$
using the following Mathematica code for the Gregory coefficients:  

`g[n_] := Integrate[x*Pochhammer[x - n + 1, n - 1]/n!, {x, 0, 1}]`

The plot below of $S_N$ versus $N$ (evaluated for $N$ up to 20) suggests convergence to 0.16653, far from the result in the OP.

<IMG SRC="https://ilorentz.org/beenakker/MO/Gregory.png"/>

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new version: the OP has a new series

$$\sum_{n=1}^\infty \sum_{k=1}^\infty\frac{|G_n|}{k}\frac{\Gamma(k+\frac{3}{2})\Gamma(n+\frac{1}{2})}{(n+k+1)!}=\frac{\pi(-5+8\log 2)}{8}=0.214091,$$

which I have also evaluated numerically and seems to converge to 0.186.