The algorithm is still exponential, it just not in the class that the ETH claims. In practice it's likely worse than current algorithms for any reasonable input you might throw at it, and starts getting faster only once the problem is already too big.

2^(n/logn) will eventually be smaller than (1+epsilon)^n for any fixed positive epsilon. However, as log(n) grows quite slowly n will have to be exceptionally large. For epsilon=2^-D n will have to be something like 2^(2^D) large.