Earlier quoted context omitted.
I'm sorry, but I'm pretty sure you are wrong there. m^(log(log(m))/log(m)) = log(m), and log(log(m))/log(m) is certainly not o(1) since it's nondecreasing (and that's just a single logarithm). I'm pretty sure o(1) requires your slowdown over linear to be faster asymptotically than any iterated logarithm, (so faster than log(log(m)), log(log(log(m))), etc.), which is close enough to linear time that calling it "almost…
log(log(m))/log(m) tends to zero as m tends to infinity, which is what it means to be o(1). It is decreasing for m > e^e. Besides n^(1+o(1)), the other common definition for "almost linear" is precisely O(n log^k (n)) for some k, no matter how large.
For an example, consider 2^sqrt(log(n)).
This is a bit similar to something being faster than polynomial, but slower than exponential.