$x_0=2,\ x_n=x_{n-1}+\log \left(x_{n-1}\right)\quad$ has a series expansion about $1.$
Since $x_0=2,$ $(x_0-1)^k=1,$ so the recurrence can be written, up to the first $5$ terms as
\begin{align} 1+\frac{773788}{604800}\ 2^n-\frac{247200}{604800}\ 2^{2 n}+\frac{118825}{604800}\ 2^{3 n}-\frac{54300}{604800}\ 2^{4 n}+\frac{13687}{604800}\ 2^{5 n}-\dots \end{align}
found with
Evaluate[Plus @@ (((FindSequenceFunction[With[{nn = #}, (Rest@(List @@
Series[Nest[# + Log[#] &, x, #], {x, 1, 20}] & /@ Range[0, 20]))
[[All, 3, nn + 1]]]] & /@ Range[0, 5] /. # -> n)[[All, 1]]))]
and then combining like terms. The problem is finding the limits of the coefficients of $2^{k n}.$ Theoretically, the recurrence relation should be rewriteable as $$1+\sum _{k=1}^{\infty } (-1)^{k+1} c_k 2^{k n}$$
but I don't know how to go about finding a general solution for $c_k.$
Added
Building on @AntonioVargas' comment below, the recurrence relation seems very close numerically to $$\sum _{k=1}^{\infty } (-1)^{k+1} \left(n c_k \log (n)-\frac{1}{2} n \log \left(c_k\right)\right)\approx \pi n/2 +n \log n$$
(based on the first $9$ terms of the series expansion).
Numerically, an approximation of the coefficients can be found with something like
Abs[With[{aa = SplitBy[SortBy[Flatten@Join[{1, 2^n}, Drop[List @@@
Expand[Evaluate[(((FindSequenceFunction[With[{nn = #}, (Rest@(List @@
Series[Nest[# + Log[#] &, x, #], {x, 1, 20(*LARGER THAN number of terms*)}]
& /@ Range[0,20(*LARGER THAN number of terms*)]))[[All, 3, nn + 1]]]] & /@
Range[0, 9(*number of terms*)] /. # -> n)[[All, 1]]))]], 2]],
Coefficient[PowerExpand[Log2[#]], n] &], Coefficient[PowerExpand[Log2[#]],
n] &]}, (a /. Solve[Plus @@ (aa[[#]]) == a 2^( (# - 1) n), a][[1]] // N) &
/@ Range[2, Length@aa]]]