Potentially stronger form of non-$ETH$
If $\alpha \in O\left(\frac{\log\log n}{ \log n}\right)$, then you would have proved that $P = NP$, since $n^{\alpha} \in O(\log n)$ and $2^{n^{\alpha}} \in O(\mathrm{poly}(n))$. 

On the other hand, for $\alpha \in \omega\left(\frac{\log\log n}{ \log n}\right)$ and $\alpha \in o(1)$, I think the most interesting thing that happens is breaking ETH with all its consequences (including breaking SETH and many other conditional lower-bounds on polynomial and parameterized problems).