It helps to rewrite the expression from Gradshteyn, $$Q_\nu^0(z)=\frac{ \Gamma \left(\frac{1}{2}\right) \Gamma (\nu+1)\, _2F_1\left(\frac{\nu}{2}+1,\frac{\nu}{2}+\frac{1}{2};\nu+\frac{3}{2};\frac{1}{z^2}\right)}{2^{\nu+1}z^{\nu+1} \Gamma \left(\nu+\frac{3}{2}\right)},$$ in terms of the regularized hypergeometric function, $$Q_\nu^0(z)=\sqrt{\pi } (2z)^{-\nu-1} \Gamma (\nu+1) \, _2\tilde{F}_1\left(\tfrac{\nu}{2}+1,\tfrac{\nu}{2}+\tfrac{1}{2};\nu+\tfrac{3}{2};\frac{1}{z^2}\right).$$ You can then set $\nu=-3/2$ and transform back to the ordinary hypergeometric function, $$Q_{-3/2}^0(z)=\frac{\pi \, _2F_1\left(\frac{3}{4},\frac{5}{4};2;\frac{1}{z^2}\right)}{4 \sqrt{2} z^{3/2}}.$$
Numerically, you can evaluate $Q^0_{-3/2}(z)$ as the real part of Mathematica's LegendreQ[-3/2,z].