A good approximation I have found for $p_{n}$ is
\begin{align} \int_{2}^{n}\log (x \log (x \log (x)))\ dx\\ \end{align}
and seems to be a better estimate than $n \log (n)$.
The error term seems to agree with the asymptotic expansion of Cipolla:
$$p_n=n\log n+n\log\log n-n+n\frac{\log\log n}{\log n}+O(n(\log\log n/\log n)^2)$$
(from this MO thread) where the $O$ term is replaced for a constant (in this case, $-e$).
Perhaps more interestingly, it seems that applying successive $\log$ terms to the integral
\begin{align} &\int_{2}^{n}\log (x)\ dx&\tag{1}\\ &\int_{2}^{n}\log (x \log (x))\ dx &\tag{2}\\ &\int_{2}^{n}\log (x \log (x \log (x)))\ dx &\tag{3}\\ &\int_{2}^{n}\log (x \log (x \dots \log (x)))\ dx &\tag{4}\\ \end{align}
seems to move very slowly towards a better asymptotic. Is this the case?
Added
Numerical results at @RghtHndSd's request:
Firstly comparing $\color{blue}{n \log (n)},\ \color{green}{\int_{2}^{n}\log (x \log (x \log (x)))\ dx}$ and $\color{red}{p_{n}}$ at successive powers of $10$
with
p[x_] := Round@(x Log[x])
p3[x_] := Round@NIntegrate[Log[n Log[n Log[n]]], {n, 2, x}]
Grid[tab2 = Table[{Style[p[n], FontColor -> Blue],
Style[p3[n], FontColor -> Darker@Green],
Style[Prime[n], FontColor -> Red]}, {n, Table[10^j, {j, 1, 12}]}],
ItemSize -> All, Alignment -> Left]
and then comparing successive $\log$ terms
with
p1[x_] := Round@NIntegrate[Log[n], {n, 2, x}]
p2[x_] := Round@NIntegrate[Log[n Log[n]], {n, 2, x}]
p3[x_] := Round@NIntegrate[Log[n Log[n Log[n]]], {n, 2, x}]
p4[x_] := Round@Re@NIntegrate[Log[n Log[n Log[n Log[n]]]], {n, 2, x}]
p5[x_] := Round@Re@NIntegrate[Log[n Log[n Log[n Log[n Log[n]]]]], {n, 2, x}]
Grid[tab2 = Table[{p1[n], p2[n], p3[n], p4[n], p5[n]},
{n, Table[10^j, {j, 1, 12}]}], ItemSize -> All, Alignment -> Left]
which, although p5[n] exceeds $p_{n}$ at $10^{12}$, the successive log terms added to the integral clearly approach a limit near to the asymptote of $p_{n}$. This is not as good, I think, as $\operatorname{li}^{-1}(n)$, but this is difficult to compute for large $n$.
Improved results
It may be that
\begin{align} &\int_{2}^{n}\log (x \log (x \log (x)/e))\ dx\\ \end{align}
is better.
\begin{align} && \text{left hand column}&\quad\left|1-n \dfrac{\log(n)}{p_{n}}\right|&\\ \\ && \text{middle column}&\quad\left|1-\dfrac{\int_{2}^{n}\log (x \log (x \log (x)))\ dx}{p_{n}}\right|&\\ \\ && \text{right hand column}&\quad\left|1-\dfrac{\int_{2}^{n}\log (x \log (x \log (x)/e))\ dx+\frac{9}{2}\sqrt{n}}{p_{n}}\right|&\\ \end{align}
running from $10^{1}$ to $10^{12}$.
Minor improvements
As daniel says, it is likely there are a lot of trailing terms. The best I have been able to manage so far is
\begin{align} &\int_{2}^{n}\log (x \log (x \log (x)/e))\ dx+4 \sqrt{x}+\frac{x}{23 \log ^2(x)}+\frac{x}{23 \log (x)}+\frac{\sqrt{x}}{23 \log (x)}\\ \end{align}
but it is only really accurate up to $10^{11}$
which can be seen to tail off just after $10^{11}$.
pbs[x_] := Re@NIntegrate[Log[n Log[n Log[E^E, n]]], {n, 2, x}]
+ 4 Sqrt[x] + Sqrt[x]/(23 Log[x]) + x/(23 Log[x]) + x/(23 Log[x]^2)
GraphicsGrid[{
Table[ListLinePlot[Transpose@Table[{(pbs[n] - Prime[n])
/(30 Sqrt[n]/Log[n])}, {n, 2, 10^k, 10^(k - 1)/2}]], {k, 2, 4}],
Table[ListLinePlot[Transpose@Table[{(pbs[n] - Prime[n])
/(30 Sqrt[n]/Log[n])}, {n, 2, 10^k, 10^(k - 1)/2}]], {k, 5, 7}],
Table[ListLinePlot[Transpose@Table[{(pbs[n] - Prime[n])
/(30 Sqrt[n]/Log[n])}, {n, 2, 10^k, 10^(k - 1)/2}]], {k, 8, 10}],
Table[ListLinePlot[Transpose@Table[{(pbs[n] - Prime[n])
/(30 Sqrt[n]/Log[n])}, {n, 2, 10^k, 10^(k - 1)/2}]], {k, 11, 12}]}, ImageSize -> 1000]]
