Can SAT be solved in time n^k, for a specific k? The P vs NP problem is open. How about the following questions--Can SAT be done in $n^k$ time for some specific $k$?
Why do I ask these questions? Ben-David and Halevi's paper On the independence of P versus NP proves that if P = NP is independent of PA, then SAT can be solved in $n^{g(n)}$ time, where $g$ is a very slow, almost constant function. This means that if we can neither prove nor disprove SAT is in P, then SAT lies on the boundary of P. It's not in P and it's not outside P either. So there's a gray area near the boundary of P. Because of this possibility, I think the P vs NP problem is not a good formulation. I therefore propose to ask more precise questions like whether SAT can be solved in linear/quadratic/cubic/etc time.
 A: It seems you are looking for lower bounds on SAT, not upper bounds. In that case, see this question I asked here a while ago. In short, the best lower bounds we have for SAT are linear, so can't even say that SAT cannot be solved in O(n) time.
Secondly I would just like to point out that Ben-David and Halevi's paper does not claim what you wrote. It says that if P vs NP is proved to be independent of PA (or ZFC) using currently known techniques then NP is contained in DTIME($n^{g(n)}$) for infinitely many inputs, where g(n) is an extremely slow growing function. Note the "infinitely many inputs" part, and most importantly, the "using currently known techniques" part.
A: As far as I know SAT is NP-Complete. Therefore, if there was such a $k$ as you said, then SAT would be in $P$, because, you know, $n^k$ is a polynomial. Thus, finding such a $k$ would prove $P=NP$.
A: You might find this paper by Patrascu and Williams interesting. It surveys the state of the art for SAT, as well as discussing implications for improved bounds. 
