The relationship between degree of vertex and size of dominating set
If all vertices have degree at least $d$ then there is always a dominating set of size $\frac{n \ln n}{d} + 1$. Pick a random set $S$ of size $\frac{n \ln n}{d}$. For a particular vertex $v$, the probability that $v$ is not dominated by $S$ is upper bounded by $(1 - \frac{d}{n})^{\frac{n\ln n}{d}} \leq e^{-\ln n} = \frac{1}{n}$. Introduce an indicator variable for every vertex that is $1$ if it is not dominated by $S$ and $0$ otherwise. The number of undominated vertices is the sum of indicator variables, so the expected number of undominated vertices is at most $n \cdot \frac{1}{n} = 1$. Hence there exists a set of size $\frac{n \ln n}{d}$ that leaves at most one vertex undominated, add this vertex to the dominating set.    

For your question it means that if all vertices have degree at least $n^k$ then there is a dominating set of size $n^{1-k}\log n + 1$. Would be nice to see if one can get rid of the $\log n$ or not.