To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is
![\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%7Bk%20%3D%202%7D%5E%7B3%7D%20%5Csqrt%5Bk%5D%7B%5Ccos%28kx%29%7D%20%7D%7B%20%7Bx%7D%5E%7B2%7D%20%7D)
Let a = 1 and b the cosine product, and write them as

with
![b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}](https://tex.z-dn.net/?f=b%20%3D%20%5Csqrt%7B%5Ccos%282x%29%7D%20%5Csqrt%5B3%5D%7B%5Ccos%283x%29%7D%20%3D%20%5Csqrt%5B6%5D%7B%5Ccos%5E3%282x%29%7D%20%5Csqrt%5B6%5D%7B%5Ccos%5E2%283x%29%7D%20%3D%20%5Cleft%28%5Ccos%5E3%282x%29%20%5Ccos%5E2%283x%29%5Cright%29%5E%7B%5Cfrac16%7D)
Now we use the identity

to rationalize the numerator. This gives

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

For the remaining limit, use the Taylor expansion for cos(x) :

where
essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.
Then



so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.
Edit: some scratch work suggests the limit is 10 for n = 6.