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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, $$Q_\nu^0(z)=\sqrt{\pi } (2z)^{-\nu-1} \Gamma (\nu+1) \, _2\tilde{F}_1\left(\tfrac{\nu}{2}+1,\tfrac{\nu}{2}+\tfrac{1}{2};\nu+\tfrac{3}{2};\frac{1}{z^2}\right).$$ This representation remains well-defined for $\nu=-3/2$. You can then transform back to the ordinary hypergeometric function, by means of the identity $$\, _2\tilde{F}_1(a,b;0;x)=a b x \, _2F_1(a+1,b+1;2;x),$$ arriving at $$Q_{-3/2}^0(z)=\frac{\pi \, _2F_1\left(\frac{5}{4},\frac{3}{4};2;\frac{1}{z^2}\right)}{4 \sqrt{2} z^{3/2}}.$$

Numerically, you can evaluate $Q^0_{-3/2}(z)$ as the real part of Mathematica's LegendreQ[-3/2,z].

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, $$Q_\nu^0(z)=\sqrt{\pi } (2z)^{-\nu-1} \Gamma (\nu+1) \, _2\tilde{F}_1\left(\tfrac{\nu}{2}+1,\tfrac{\nu}{2}+\tfrac{1}{2};\nu+\tfrac{3}{2};\frac{1}{z^2}\right).$$ This representation remains well-defined for $\nu=-3/2$. You can then transform back to the ordinary hypergeometric function, by means of the identity $$\, _2\tilde{F}_1(a,b;0;x)=a b x \, _2F_1(a+1,b+1;2;x),$$ arriving at $$Q_{-3/2}^0(z)=\frac{\pi \, _2F_1\left(\frac{5}{4},\frac{3}{4};2;\frac{1}{z^2}\right)}{4 \sqrt{2} z^{3/2}}.$$

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].

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, $$Q_\nu^0(z)=\sqrt{\pi } (2z)^{-\nu-1} \Gamma (\nu+1) \, _2\tilde{F}_1\left(\tfrac{\nu}{2}+1,\tfrac{\nu}{2}+\tfrac{1}{2};\nu+\tfrac{3}{2};\frac{1}{z^2}\right).$$ This representation remains well-defined for $\nu=-3/2$. You can then transform back to the ordinary hypergeometric function, by means of the identity $$\, _2\tilde{F}_1(a,b;0;x)=a b x \, _2F_1(a+1,b+1;2;x),$$ arriving at $$Q_{-3/2}^0(z)=\frac{\pi \, _2F_1\left(\frac{5}{4},\frac{3}{4};2;\frac{1}{z^2}\right)}{4 \sqrt{2} z^{3/2}}.$$

Numerically, you can evaluate $Q^0_{-3/2}(z)$ as the real part of Mathematica's LegendreQ[-3/2,z].

You can take the limit $\nu\rightarrow -3/2$ of the expression from Gradshteyn $$Q_\nu^0(z)=\frac{ \Gamma \left(\frac{1}{2}\right) \Gamma (\nu+1)\, _2F_1\left(\frac{\nu}{2}+1,\frac{\nu}{2}+\frac{1}{2};\nu+\frac{3}{2};\frac{1}{z^2}\right)}{2^{\nu+1}z^{\nu+1} \Gamma \left(\nu+\frac{3}{2}\right)},$$ which gives $$Q_{-3/2}^0(z)=\frac{\pi \, _2F_1\left(\frac{3}{4},\frac{5}{4};2;\frac{1}{z^2}\right)}{4 \sqrt{2} z^{3/2}}.$$

Alternatively, you can evaluate $Q^0_{-3/2}(z)$ as the real part of Mathematica's LegendreQ[-3/2,z].

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].