Is there an explicit formula that gives the value of $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$ for $n$ square roots?

Elaborating on Michael Rozenberg's answer:

Note that

$$\sqrt{2+2\cos\alpha} = \sqrt{4\cos^2\left(\frac{\alpha}{2}\right)} = 2\cos\left(\frac{\alpha}{2}\right)$$

So,

$$\sqrt{2} = 2\cos\left(\frac{\pi}{4}\right)$$

$$\sqrt{2+\sqrt{2}} = 2\cos\left(\frac{\pi}{8}\right)$$

$$\vdots$$

Thus, if we have $n$ square roots, we have

$$x=2\cos\left(\frac{\pi}{2^{n+1}}\right)$$

Just take $x=2\cos\alpha$ and the rest is smooth.

In general, the number $\sqrt{2+\sqrt{2+\sqrt{\dots+\sqrt{2}}}}$ consisting of $n$ radices is algebraic of degree $2^n$ and has conjugate roots $\pm\sqrt{2\pm\sqrt{2\pm\sqrt{\dots\pm\sqrt{2}}}}$. This could in theory give you its minimal polynomial, with the help of Vieta's formulas.

Example for the case $n=2$ goes as follows. The roots are $$ x_1=\sqrt{2+\sqrt2},\quad x_2=-\sqrt{2+\sqrt2},\quad x_3=\sqrt{2-\sqrt2},\quad x_4=-\sqrt{2-\sqrt2}. $$ The product of all the roots is $x_1x_2x_3x_4 = -(2+\sqrt2)\times-(2-\sqrt2) = 4-2=2$, sum is $x_1+x_2+x_3+x_4=0$, and the other necessary symmetric functions are $x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=(x_1+x_2)(x_3+x_4)+x_1x_2+x_3x_4=-(2+\sqrt2)-(2-\sqrt2)=-4$ and $x_1x_2x_3+x_1x_2x_3+x_1x_3x_4+x_2x_3x_4=x_1x_2(x_3+x_4)+(x_1+x_2)x_3x_4=0$.

Therefore we get that the polynomial is $x^4-4x^2+2$. There is not much more to say about the numbers.