Arbitrary Turing machine run time analysis on the empty word

Consider $$L = \{ \langle M,n \rangle : M$$ accpets $$\epsilon$$ in less than $$T(n)$$ steps$$\}$$

This language is decidable because a decider can simulate $$M$$ on $$\epsilon$$ and accept if it accepts and reject if it rejects or passed more than $$T(n)$$ steps.

The decider above will run in $$O(T(n)log(T(n)))$$ time corresponding to simulation time bounds (maybe even $$O(T(n))$$ has been achieved for simulation time of $$T(n)$$ steps)

but since we are dealing with a constant word, might it have a better simulation time? $$o(T(n))$$?

The $$\log T(n)$$ factor in the analysis stems from the need to keep (and update) a step counter, in binary, that goes up to $$T(n)$$. This requires $$\log(T(n))$$ cells, and the need to update it (or rather, to drag it along on the tape) costs $$\log(T(n))$$ in every iteration.
Now, if you could avoid this for $$\epsilon$$, then you could avoid it for an arbitrary input word $$w$$, by modifying the given TM $$M$$ to a new TM $$M'$$ that first writes $$w$$ on the tape, and then simulates $$M'$$ for $$T(n)+|w|$$ steps on $$\epsilon$$.
Since $$T(n)$$ is super-linear, this would contradict the best known lower bounds.
Note that this does not prove that a better simulation is not possible, only that the word $$\epsilon$$ plays no special role.
In fact, however, you could prove that $$o(T(n))$$ is not achievable, by tweaking slightly the Time Hierarchy Theorem