How many values does $\sqrt{\sqrt{i}}$ have?

Your list of four solutions only has two - as has been pointed out you listed two of them twice. (In the original post, anyway...) But there are four roots.

First, $i=e^{i\pi/2}=e^{i5\pi/2}$ leads gives two values for $\sqrt i$, namely $e^{i\pi/4}$ and $e^{i5\pi/4}$. Each of these has two square roots: Since $e^{i\pi/4}=e^{i9\pi/4}$ it has the two square roots $e^{i\pi/8}$ and $e^{i9\pi/8}$. Similarly $e^{i5\pi/4}=e^{i13\pi/4}$ has the two square roots $e^{i5\pi/8}$ and $e^{i13\pi/8}$. For a total of four values of $\sqrt{\sqrt i}$, namely $$e^{i\pi/8},e^{i5\pi/8},e^{i9\pi/8},e^{i13\pi/8}.$$

NOTE Others have suggested that although a complex number has two square roots, the notation $\sqrt z$ refers to only one of them. I think it's wrong to put it that way; that notation refers to only one of them if we have clearly stated in advance which "branch" we're referring to. But Wolfram is wrong in any case. By my lights, $i$ has two square roots, each of which has two square roots, for a total of four as above. But if we are saying that the notation $\sqrt z$ refers to one square root, then Wolfram should say there's only one value for $\sqrt{\sqrt i}$, namely the one square root of the one square root.

I think four is the right number. With that convention there's only one; there's no sensible way to give exactly two values, as WA does. You can't believe everything they tell you.

Trick is to introduce an unknown, $z$. Write $z=\sqrt{ \sqrt{i}}$. We want to know how many values are possible for $z$. Squaring we get $z^2=\surd i$. Again squaring we get $z^4-i=0$. This is a polynomial of degree 4 over the complex numbers. So it has 4 roots. Taking the derivative shows that the roots are distinct.

SO we have 4 distinct values for the number.

EDIT (to clarify the doubts raised by SHailesh and Piquito in the comments):

For a polynomial $f(z)$ to have $\alpha$ as repeated root means it is of the form $(z-\alpha)^2 g(z)$ for some other polynomial $g(z)$.Calculating derivative of $f(z)$ using product rule makes it clear that $\alpha$ will also be a root of the derivative $f'(z)$; and converse is also true.

@Piquito: Fundamental theorem of algebra states that the number of roots is equal to the degree of the polynomial only when we count them with multiplicity and include complex roots, real irrational roots.