Let $u = e^{ix}$. Then $\cos x \cos 2x ... \cos nx = \frac{1}{2^n}\prod_{k=1}^n (u^k + u^{-k})$.
Write $\prod_{k=1}^n (z^k + z^{-k}) = \sum_{j=-N}^N a_j z^j$
where $N=\frac{n(n+1)}2$.
Then the integral above is just $\frac{2\pi}{2^n}a_0$.
Combinatorially, we can see that $a_0$ is the number of ways of picking a subset of $\{1,2,...,n\}$ which adds up to $\frac{n(n+1)}{4}$.
Now, the set of subsets of $\{1,...,n\}$ that add up to a particular value is an antichain, so we have a bound on the size of $a_0$, namely, $0\leq a_0 \leq \binom {n}{\lfloor n/2\rfloor}$.
And we know that $\lim_{n\to\infty}\frac{1}{2^n}\binom {n}{\lfloor n/2\rfloor} =0$
So we know your limit is zero.
(The maximum size of an antichain in the Boolean poset is Sperner's theorem.)
Note that when $n\equiv 1,2\pmod 4$, $\frac{n(n+1)}4$ is not an integer, so in those cases, $a_0=0$. Anybody have an "interesting" lower bound on $a_0$ for the other cases?
[In comments below, Robert Israel shows that when $n\equiv 0,3\pmod 4$, $a_0\geq 2^{\lfloor (n+1)/4\rfloor}$. I suspect that there is a polynomial, $q$, such that $a_0\geq \frac{2^n}{q(n)}$.]
This proof generalizes. Given any sequence of positive integers, $\{b_1,...,b_n\}$, we have the same result:
$\int_{0}^{2\pi} \cos b_1 x \cos b_2 x ... \cos b_n x dx= \frac{2\pi}{2^n}A$
where $A$ is the count of an antichain, so $0\leq A\leq \binom {n}{\lfloor n/2\rfloor}$
If $b_i=1$ for all $i$, then we get exact equality when $n$ is even, which is to say that:
$\int_{0}^{2\pi} \cos^{2m} x dx = \frac{2\pi}{2^{2m}} \binom {2m}{m}$