Trigonometric functions have got large powers!

First, we try to simplify the given expression. (Although, you can substitute back the values of $\sin \frac{\pi}{6}$ and $\cos \frac{\pi}{6}$ to make your work easier) but I'll present a general solution.

Multiplying the expression by $\cos^2x-\sin^2x$ , to simplify it as follows :

$\begin{aligned} F_n(x)(\cos^2x-\sin^2x) &= & (\cos^2 x-\sin^2 x) \prod_{k=1}^n (\cos^{2^n}x + \sin^{2^n}x) \\ F_n(x)\cos 2x & = & (\cos^2 x-\sin^2 x)(\cos^2 x+\sin^2 x)(\cos^4 x+\sin^4 x) \ldots (\cos^{2^n} x+\sin^{2^n} x) \\ & = & (\cos^4 x-\sin^4 x)(\cos^4 x+\sin^4 x) \ldots (\cos^{2^n} x+\sin^{2^n} x) \\ & = & (\cos^8 x-\sin^8 x)(\cos^8 x+\sin^8 x) \ldots (\cos^{2^n} x+\sin^{2^n} x) \\ & & . \\ & & . \\ & & . \\ & = & (\cos^{2^{n+1}} x - \sin^{2^{n+1}} x) \\ \end{aligned}$

Thus, $F_n(x) = \frac{\cos^{2^{n+1}} x -\sin^{2^{n+1}} x}{\cos 2x}$

We need to compute $F_{2015}\left(\frac{\pi}{6}\right)$

$\begin{aligned} F_{2015}\left(\frac{\pi}{6}\right) & = & \frac{\cos^{2^{2016}} \left(\frac{\pi}{6}\right) -\sin^{2^{2016}} \left(\frac{\pi}{6}\right)}{\cos \left(\frac{\pi}{3}\right)} \\ & = & \left(\frac{\sqrt{3}}{2}\right)^{2^{2016}} - \left(\frac{1}{2}\right)^{2^{2016}} \times \frac{1}{\frac{1}{2}} \\ & = & \frac{\sqrt{3}^{2^{2016}}-1}{2^{2^{2016}-1}} \\ & = & \frac{3^{2^{2016-1}}-1}{2^{2^{2016}-1}} \end{aligned}$

And thus, $a=3,\ b=2016,\ c=2 \Rightarrow a+b+c = \boxed{2021}$

Right off the bat, we know that $\sin x = \frac{1}{2}$ and $\cos x=\frac{\sqrt{3}}{2}$ so we can simplify the product to $F_{2015}(x) = \prod_{k=1}^{2015} \big( \left(\frac{1}{2}\right)^{2^k} +\left(\frac{3^{\frac{1}{2}}}{2}\right)^{2^k} \big)$ $= \prod_{k=1}^{2015} \left(\frac{1+3^{2^{k-1}}}{2^{2^k}}\right) = \dfrac{\prod_{k=1}^{2015} \left(1+3^{2^{k-1}}\right)}{\prod_{k=1}^{2015} \left(2^{2^k}\right)}$

We continue by evaluating the top and the bottom individually. For the denominator, we have $\prod_{k=1}^{2015} \left(2^{2^k}\right) = 2^{\sum_{k=1}^{2015}2^k} = 2^{2^{2016}-2}$

And for the numerator, we have $\prod_{k=1}^{2015} \left(1+3^{2^{k-1}}\right) = \dfrac{3^{2^{2015}}-1}{2}$ which follows from the fact that $\prod_{k=1}^{q} \left(1+a^{2^{k-1}}\right) = \dfrac{a^{2^q}-1}{a-1}$ which can be proven easily by induction. Putting these results together, we get $\dfrac{\prod_{k=1}^{2015} \left(1+3^{2^{k-1}}\right)}{\prod_{k=1}^{2015} \left(2^{2^k}\right)} = \dfrac{\dfrac{3^{2^{2015}}-1}{2}}{2^{2^{2016}-2}} = \dfrac{3^{2^{2015}}-1}{{2^{2^{2016}-1}}}.$

Now for the hard part: our final answer is $3 + 2016 + 2 = \boxed{2021}$ .