Still not a complete answer, but a method to be completed or improved.

Denote $f_n(t)=\frac{2 - e^{-t}\left ( 2 + 2t+t^2 \right )}{t^3}\left ( \frac{1-e^{-t}}{t} \right )^{n-1}$. Assume that we have found the coefficients $c_1,c_2,\dots,c_n$ and polynomials $g_1,\dots,g_n$ such that $$f_n(t)=2t^{-n-2}+\sum_{k=1}^n(g_k(t^{-1})e^{-kt})'+\sum_{k=1}^n c_k \frac{e^{-kt}}t.\,\,(*)$$
Then $\sum c_k=0$ (else $f_n$ would have a non-zero residue at 0, which is absurd). We have $\int_0^\infty \sum_{k=1}^n c_k \frac{e^{-kt}}t dt=-\sum c_k\log k$ by [Frullani integrals][1].
The integral of the other part of our sum $(*)$ is minus the limit at zero of the function $-\frac2{n+1}t^{-n-1}+\sum g_k(t^{-1})e^{-kt}$. The limit must exist, since the initial integral converges. 

Now I cheat a bit. Note that if we write $f_n(t)=\sum_{k=0}^n q_k(1/t) e^{-kt}$ for polynomials $q_k$, then $c_k$ equals to the residue of the $q_k(1/t) e^{-kt}$, which is pretty computable. If I am not mistaken, $$q_k(t^{-1})=(-1)^kt^{-n-2}\left(2\binom{n-1}k+\binom{n-1}{k-1}(2+2t+t^2)\right)=\\=(-1)^kt^{-n-2}\left(2\binom{n}k+\binom{n-1}{k-1}(2t+t^2)\right).$$ Thus $$c_k=2(-1)^{k+n+1}\binom{n}k\frac{k^{n+1}}{(n+1)!}+2(-1)^{k+n}\binom{n-1}{k-1}\frac{k^{n}}{n!}+(-1)^{k+n+1}\binom{n-1}{k-1}\frac{k^{n-1}}{(n-1)!}=\\
=(-1)^{k+n+1}\frac{k^{n-1}(n^2+n-2k)}{n(n+1)(n-k)!(k-1)!}.$$
This matches a coefficient of $\log 3$ for $n=4$ from Shahrooz Janbaz's answer, you may check others for be sure. 
It remains to prove Sylvain JULIEN's guess for the rational part. 

  [1]: http://mathworld.wolfram.com/FrullanisIntegral.html