Approximating the inverse logarithmic integral Numerical evidence suggests that
\begin{align}
&\operatorname{li}^{-1}(n)=x_n+\mathcal{O}\left(\log^2 n\right)\\
\end{align}
where 
\begin{align}
&x_n=x_{n-1}+\log \left(x_{n-1}\right)\\
\end{align}
and $\operatorname{li}^{-1}$ is the inverse logarithmic integral function.
Transpose[{(InverseFunction[LogIntegral][N[#]]), 
(Nest[# + Log[#] &, N[Pi/2], #] + Divide[(Log@#)^2, 10.5] + 7.5)} 
& /@ Range[3, 1000]] // ListLinePlot

It this the case? If so, can the relationship be described more precisely?
 A: $\DeclareMathOperator{\li}{li}$The claim is correct and can be improved.
Theorem: Consider any primitive $\li:\,]1,\infty[\mathbb{R}$ of $1/\log(x)$ and 
any sequence $x_n$, $n=0,1,\dots$, of real numbers satisfying $x_0>1$ and $x_n=x_{n-1}+\log(x_{n-1})$ for $n\geq1$.
Then there is a constant $C$ such that 
$$ \li(x_n)+\tfrac12\log(\log(x_n))=n+C+O\left(\frac1{n\log(n)}\right)\mbox{ as }n\to\infty.$$
As a consequence, with the above constant,
$$x_n=\li^{-1}(n)+\log(n\log(n))\cdot(C-\tfrac12\log(\log(n)))+o(1)$$
and, in particular, $x_n=\li^{-1}(n)+O(\log(n)\log(\log(n)))$ for large $n$. 
Proof: Clearly, the sequence $x_n$ is stricly increasing and tending to $\infty$, even $x_n/n$ tends to infinity.
As for other recurrence relations, it is useful to have a function $F$ with
$$F(x+\log(x))\approx F(x)+1\mbox{ for large }x$$
because then $F(x_{n+m})\approx F(x_m)+n$. Solving approximately $F'(x)\log(x)+\frac12F''(x)\log(x)^2\approx 1$, we find
$F'(x)\approx \frac 1{\log(x)}+\frac1{2x\log(x)}$.
We first use $F(x)=\li(x)$ and find that $F(x+\log(x))-F(x)=\frac{\log(x)}{\log(x+d(x))}$ with some $d(x)$ between $0$
and $\log(x)$. Hence $F(x+\log(x))-F(x)\to1$ as $x\to+\infty$ and $F(x_{n+1})-F(x_n)\to 1$ as $n\to\infty$. By Cesaro's theorem
we obtain that also 
$$\frac1n(F(x_n)-F(x_0))=\frac1n((F(x_n)-F(x_{n-1})+\cdots+(F(x_1)-F(x_0)))\to1.$$
This means that $\li(x_n)\sim n$ as $n\to\infty$. Since $\li(x)\sim \frac x{\log(x)}$ as $x\to+\infty$ by L'Hôpital's rule
this implies that $\frac{x_n}{\log(x_n)}\sim n$. Estimating the unique zero of $x-\alpha\log(x)$ in $]e,\infty[$ for large $\alpha$,
this implies that $x_n\sim n\log(n)$ as $n\to\infty$.
Now define $F(x)=\li(x)+\frac12\log(\log(x))$ for $x>e$. Then a small calculation using Taylor's formula yields
$F(x+\log(x))-F(x)=1+O\left(\frac{\log (x)}{x^2}\right)$ for $x>1$.
Hence $F(x_{n+1})=F(x_n)+1+O\left(\frac{\log (x_n)}{x_n^2}\right)$ for all $n$.
Since $x_n\sim n\log(n)$ we have $F(x_{n+1})=F(x_n)+1+O\left(\frac1{n^2\log(n)}\right).$
Hence $F(x_n)-n$ is a Cauchy sequence and tends to some constant we call $C$. We find
$$F(x_n)-n-C=O\left(\sum_{k=n}^\infty\frac1{k^2\log(k)}\right)=O\left(\frac1{n\log(n)}\right).$$ 
This proves the first statement of the Theorem.
For the second statement, we first write $x_n=\li^{-1}\left(n+C-\frac12\log(\log(x_n))+O\left(\frac1{n\log(n)}\right)\right)$.
Then we use a Taylor expansion around $n$ and that $x_n\sim n\log(n)$, $\li^{-1}(n)= n(\log(n)+O(\log(\log(n)))$
$\left(\li^{-1}\right)'(n)=\log\left(\li^{-1}(n)\right)= \log(n\log(n))+O\left(\frac{\log(\log(n))}{\log(n)}\right)$ and 
$\left(\li^{-1}\right)''(n)=\frac{1}{\li^{-1}(n)}\log\left(\li^{-1}(n)\right)\sim \frac1n$ and find the 
wanted approximation.
Edit: In the same way, a third term in $F$ can be found and we obtain an improvement of
the first statement of the theorem. With $F(x)=\li(x)+\frac12\log(\log(x))-\frac1{12x}-
\frac1{12}\li(\frac1x)$, we find $F(x+\log(x))-F(x)=1+O\left(\frac{\log (x)}{x^3}\right)$.
As above, this yields 
$$\li(x_n)+\frac12\log(\log(x_n))-\frac1{12x_n}-
\frac1{12}\li\left(\frac1{x_n}\right)=n+C+O\left(\frac1{n^2\log(n)^2}\right).$$
