If one were to show that an NP-complete problem had $2^{n^{O(1)/\log{\log{n}}}}$ circuit complexity, what would the consequences of this be?
NP-complete problems and sub-expenential sized circuits
complexity theory;circuits
If such an algorithm were found, it would violate the exponential time hypothesis.  See also, for example, Are there subexponential-time algorithms for NP-complete problems?.And of course, if this were the best possible complexity for any algorithm for this problem, you would have shown that $P \ne NP$ (and in fact that $NP \ne P/\text{poly}$).  When you wrote that the problem has such-and-such circuit complexity, I don't know whether you meant (a) there exists an algorithm with that complexity, or (b) the problem has that complexity (which means there exists an algorithm of that complexity, and there is no better algorithm); this paragraph only applies if you meant (b).Beyond that, I don't think there would be any other obvious consequences.  In particular, I don't know of any significant consequences if the ETH is violated (other than the obvious one, that the theorems of the form if the ETH holds then... become moot).