Improving an approximation for the inverse of the Riemann–Siegel θ-function
We start with the asymptotics $$ \theta (t) = \frac{t}{2}\log \frac{t}{{2\pi }} - \frac{t}{2} - \frac{\pi }{8} + \frac{1}{{48t}} + \mathcal{O}\!\left( {\frac{1}{{t^3 }}} \right), $$ i.e., $$ \frac{{\theta (t)}}{\pi } + \frac{1}{8} = \frac{t}{{2\pi }}\log \frac{t}{{2\pi }} - \frac{t}{{2\pi }} + \frac{1}{{48\pi t}} + \mathcal{O}\!\left( {\frac{1}{{t^3 }}} \right). $$ This may be re-written in the form $$ \frac{{\theta (t)}}{\pi } + \frac{1}{8} = \left( {\frac{t}{{2\pi }} + g(t)} \right)\log \left( {\frac{t}{{2\pi }} + g(t)} \right) - \left( {\frac{t}{{2\pi }} + g(t)} \right), $$ where $$ g(t) = \frac{1}{{48\pi t\log \frac{t}{{2\pi }}}} + \mathcal{O}\!\left( {\frac{1}{{t^3 \log t}}} \right). $$ Thus, $$ \frac{1}{e}\left( {\frac{{\theta (t)}}{\pi } + \frac{1}{8}} \right) = \frac{{\frac{t}{{2\pi }} + g(t)}}{e}\log \frac{{\frac{t}{{2\pi }} + g(t)}}{e}, $$ i.e., $$ \frac{{\frac{{\theta (t)}}{\pi } + \frac{1}{8}}}{{W\!\left( {\frac{1}{e}\left( {\frac{{\theta (t)}}{\pi } + \frac{1}{8}} \right)} \right)}} = \frac{t}{{2\pi }} +g(t)= \frac{t}{{2\pi }} + \frac{1}{{48\pi t\log \frac{t}{{2\pi }}}} + \mathcal{O}\!\left( {\frac{1}{{t^3 \log t}}} \right). $$ Iterating this once yields $$ \frac{{\frac{{\theta (t)}}{\pi } + \frac{1}{8}}}{{W\!\left( {\frac{1}{e}\left( {\frac{{\theta (t)}}{\pi } + \frac{1}{8}} \right)} \right)}} = \frac{t}{{2\pi }} + \frac{1}{{96\pi ^2 \left[ {\frac{{\frac{{\theta (t)}}{\pi } + \frac{1}{8}}}{{W\left( {\frac{1}{e}\left( {\frac{{\theta (t)}}{\pi } + \frac{1}{8}} \right)} \right)}}} \right]\log \left[ {\frac{{\frac{{\theta (t)}}{\pi } + \frac{1}{8}}}{{W \left( {\frac{1}{e}\left( {\frac{{\theta (t)}}{\pi } + \frac{1}{8}} \right)} \right)}}} \right]}} \\ + \mathcal{O}\!\left( {\frac{{\log ^2 \theta (t)}}{{\theta ^3 (t)}}} \right). $$ By solving for $t$, simplifying and introducing the inverse function, we find $$ \theta ^{ - 1} (t) = \frac{{8t + \pi }}{{4W\!\left( {\frac{{8t + \pi }}{{8\pi e}}} \right)}} - \frac{{W\!\left( {\frac{{8t + \pi }}{{8\pi e}}} \right)}}{{6 (8t + \pi )\left( {\log \left( {\frac{{8t + \pi }}{{8\pi }}} \right) - \log W\!\left( {\frac{{8t + \pi }}{{8\pi e}}} \right)} \right)}} + \mathcal{O}\!\left( {\frac{{\log ^2 t}}{{t^3 }}} \right). $$ For $t=100$ this, without the error term, gives $108.5639773824\ldots$ whereas the exact value is $108.5639773815\ldots$. It is possible to obtain higher terms by using more terms from the asymptotics of $\theta(t)$, obtaining more terms for $g(t)$ and so on. But this leads to elaborate computations once one starts iterating.
$$\theta^{-1}(x)=\frac{8 x+\pi }{4 W\left(\frac{8 x+\pi }{8 e \pi }\right)}-\frac 1{8}\left(\frac{8 x+\pi }{4 W\left(\frac{8 x+\pi }{8 e \pi }\right)} \right)^{-3/2}$$ seems to be a slight improvement $$\left( \begin{array}{cccc} x & \text{first approximation}& \text{second approximation} & \text{exact}\\ 1 & 19.67670118 & 19.67526905 & 19.67484567 \\ 2 & 21.36685143 & 21.36558582 & 21.36525782 \\ 3 & 22.95388274 & 22.95274610 & 22.95248141 \\ 4 & 24.46021637 & 24.45918309 & 24.45896286 \\ 5 & 25.90107407 & 25.90012579 & 25.89993815 \\ 6 & 27.28736031 & 27.28648338 & 27.28632040 \\ 7 & 28.62720976 & 28.62639366 & 28.62624986 \\ 8 & 29.92688609 & 29.92612257 & 29.92599401 \\ 9 & 31.19133680 & 31.19061924 & 31.19050300 \\ 10 & 32.42455244 & 32.42387543 & 32.42376931 \\ 20 & 43.56093755 & 43.56050278 & 43.56044353 \\ 30 & 53.35930910 & 53.35898840 & 53.35894405 \\ 40 & 62.37144533 & 62.37119157 & 62.37115427 \\ 50 & 70.84503043 & 70.84482081 & 70.84478766 \\ 60 & 78.91754646 & 78.91736816 & 78.91733781 \\ 70 & 86.67507580 & 86.67492089 & 86.67489261 \\ 80 & 94.17593155 & 94.17579478 & 94.17576813 \\ 90 & 101.4618807 & 101.4617584 & 101.4617331 \\ 100 & 108.5641121 & 108.5640016 & 108.5639774 \end{array} \right)$$
(this is not an answer but too long for a comment)
(+1) Interesting discussion and answers! Three years earlier I searched the best constant $C$ in following approximate value of the imaginary part of the $n$-th non-trivial zero (from your initial expression of course) : $$\;t_n\approx 2\pi\,\exp(W((n-7/8-C)/e)+1)=2\pi\dfrac{n-7/8-C}{W((n-7/8-C)/e)}$$ and conjectured that $C$ had to be exactly $\dfrac 12$ (computing different moving averages and so on). Further the actual error doesn't exceed $\pm 1$ for the first $2$ million zeros as illustrated :
Notice the vertical symmetry around $0$ and the slow decrease of the variance of the error with $n$ (a correction term depending of $n$ appears less interesting than in your question, if needed at all, since the mean error remains near $0$ for values as large as $10^{22}$ using Andrew Odlyzko's tables ).
Anyway I found this a neat illustration of the gentle statistical distribution of the zeros.
We seem further able to find the position of the $n$-th zero for $n$ as large as we want with an error of less than one (the error for the $10^4$ zeros following $10^{22}$ is less than $0.21$).
For $\,n=10^{22}+1\,$ for example the formula gives us
$t_n\approx 1370919909931995308226.770224\ $ while the actual zero is at :
$t_n= 1370919909931995308226.680161\cdots$