1. Write the complex numbers explicitly

Since all three numbers lie on the unit circle, we can write $z_k=e^{i\theta_k}$ with the given arguments: $\theta_1=-\frac{\pi}{4},\quad \theta_2=0,\quad \theta_3=\frac{\pi}{4}.$

Hence, $z_1=e^{-i\pi/4},\quad z_2=1,\quad z_3=e^{i\pi/4}.$

Also, for unit complex numbers, $\overline{z_k}=e^{-i\theta_k}.$

So, $\overline{z_2}=1,\quad \overline{z_3}=e^{-i\pi/4},\quad \overline{z_1}=e^{i\pi/4}.$

2. Compute each term

We need $z_1\overline{z_2}+z_2\overline{z_3}+z_3\overline{z_1}.$

Now, $z_1\overline{z_2}=e^{-i\pi/4}\cdot 1=e^{-i\pi/4},$ $z_2\overline{z_3}=1\cdot e^{-i\pi/4}=e^{-i\pi/4},$ $z_3\overline{z_1}=e^{i\pi/4}\cdot e^{i\pi/4}=e^{i\pi/2}=i.$

Therefore, $z_1\overline{z}_2+z_2\overline{z}_3+z_3\overline{z}_1=2e^{-i\pi/4}+i.$

Using $e^{-i\pi/4}=\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}},$ we get $2e^{-i\pi/4}=\sqrt{2}-i\sqrt{2}.$

Thus the expression becomes $\sqrt{2}+i(1-\sqrt{2}).$

3. Find its modulus squared

If $w=\sqrt{2}+i(1-\sqrt{2}),$ then $|w|^2=(\sqrt{2})^2+(1-\sqrt{2})^2.$

Now, $(\sqrt{2})^2=2,$ and $(1-\sqrt{2})^2=1+2-2\sqrt{2}=3-2\sqrt{2}.$

So, $|w|^2=2+(3-2\sqrt{2})=5-2\sqrt{2}.$

Comparing with $\alpha+\beta\sqrt{2},$ we get $\alpha=5,\quad \beta=-2.$

4. Compute $\alpha^2+\beta^2$

$\alpha^2+\beta^2=5^2+(-2)^2=25+4=29.$

5. Check options

The correct option is: $\boxed{\text{B: }29}$