Not an usual Trigonometric Summation!

We have (By-multiplying Numerator and Denominator by $\sin\left(\dfrac{x}{2^n}\right)$ ):

$\lim_{n \to \infty} \cos \left(\dfrac{x}{2} \right)\cos \left(\dfrac{x}{2^2} \right)\cos\left(\dfrac{x}{2^3} \right) \ldots \cos \left(\dfrac{x}{2^n} \right) = \dfrac{\sin(x)}{x}$

Differentiating above, we obtain:

$\sum_{n=1}^{\infty} \dfrac{1}{2^n} \tan\left(\dfrac{x}{2^n}\right)=\dfrac1x -\cot(x)$

We'll apply the above result in a step below. Now,

$L = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{3\cdot2^{k+1}}\right) = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{6} \cdot \dfrac{1}{2^k}\right)$

$= \lim_{n \to \infty} \sum_{n=1}^{\infty} \dfrac{1}{2^n} \tan\left(\dfrac{x}{2^n}\right)$ where $x=\dfrac{\pi}{6}$ .

$=\dfrac1x - \cot(x) = \dfrac{6}{\pi} - \cot\left(\dfrac{\pi}{6}\right)$

$=\dfrac{6}{\pi} - \sqrt{3}$

So, $A=6, B=1, C=3, A+B+C=\boxed{10}$ .