We want to solve for $x$ the equation
$$\psi ^{(0)}(x)=a$$
Just four months before your question, this paper gave tight bounds
$$x_{\text{min}}=\frac{1}{\log \left(1+e^{-x}\right)} \lt \psi^{-1} (x) \lt e^{x}+\frac{1}{2}=x_{\text{max}}$$
What I recently found very interesting is that is is much better to consider instead the equation
$$e^{\psi ^{(0)}(x)}=e^a$$ the lhs being very close to linearity.
Use the series expansion around $x=\alpha$
$$e^{\psi ^{(0)}(x)}=e^{\psi ^{(0)}(\alpha )}\left(1+\sum_{n=1}^\infty \frac{A_n}{n!}\,(x-\alpha)^n \right)$$ where
$$A_1=\psi ^{(1)}(\alpha )$$
$$A_2=\psi ^{(1)}(\alpha )^2+\psi ^{(2)}(\alpha )$$
$$A_3=\psi ^{(1)}(\alpha )^3+3 \psi ^{(2)}(\alpha ) \psi ^{(1)}(\alpha
)+\psi ^{(3)}(\alpha )$$ and so forth.
Truncate to some order $p$ and use power series reversion to get
$$x=\alpha +t+\sum_{n=2}^p B_n\, t^n \qquad \text{where} \qquad t=\frac{e^{a-\psi ^{(0)}(\alpha )}-1}{\psi ^{(1)}(\alpha )}$$ the very first coefficients being
$$B_2=\frac{-\psi ^{(1)}(\alpha )^2-\psi ^{(2)}(\alpha )}{2 \psi
^{(1)}(\alpha )}$$
$$B_3=\frac{2 \psi ^{(1)}(\alpha )^4+3 \psi ^{(2)}(\alpha ) \psi
^{(1)}(\alpha )^2-\psi ^{(3)}(\alpha ) \psi ^{(1)}(\alpha )+3
\psi ^{(2)}(\alpha )^2}{6 \psi ^{(1)}(\alpha )^2}$$
and so forth.
Trying for $a=\pi$ for different orders of expansion starting at the midpoint
$$\left(
\begin{array}{cc}
p & x_{(p)} \\
1 & 23.6388933011739 \\
2 & 23.6388924116017 \\
3 & 23.6388924311562 \\
4 & 23.6388924307264 \\
5 & 23.6388924307358 \\
6 & 23.6388924307356 \\
\end{array}
\right)$$
