$\def\erf{\text{erf}}$
Browsing Wolfram Functions, we found the $n$th derivative of the inverse error function and of inverse gamma regularized. Notice the regularized$\,_2\text F_2$ hypergeometric function and the Kronecker delta function $\delta_{n},\delta_{n_1,n_2}$:
$$\frac{d^n\erf^{-1}(z)}{dz^n}=\erf^{-1}(z)\delta_n+\left(\frac{\sqrt\pi}2e^{\erf^{-1}(z)^2}\right)^n\sum_{j_2=0}^n\cdots\sum_{j_n=0}^n (-1)^{\sum\limits_{k=2}^n j_k}\delta_{\sum\limits_{k=2}^n(k-1)j_k,n-1}\left(n-1+\sum_{k=2}^n j_k\right)!\prod_{k=2}^n \frac1{j_k!}\left(\frac{\sqrt\pi2^{k-1}e^{\erf^{-1}(z)^2}}{\erf^{-1}(z)^{k-1}k!}\right)^{j_k}\,_2\tilde{\text F}_2\left(\frac12,1;1-\frac k2,\frac{3-k}2;-\erf^{-1}(z)^2\right)^{j_k}\tag1$$ with the Mathematica code:
D[InverseErf[z], {z, n}] == KroneckerDelta[n] InverseErf[z] + (Pi^(n/2)/2^n) E^(n InverseErf[z]^2) Sum[\[Ellipsis] Sum[KroneckerDelta[Sum[(i - 1) Subscript[j, i], {i, 2, n}], n - 1] (-1)^Sum[Subscript[j, i], {i, 2, n}] (n - 1 + Sum[Subscript[j, i], {i, 2, n}])! Product[(1/Subscript[j, i]!) ((2^(-1 + i) E^InverseErf[z]^2 Sqrt[Pi] InverseErf[z]^(1 - i))/i!)^Subscript[j, i] HypergeometricPFQRegularizedq[{1/2, 1}, {1 - i/2, (3 - i)/2}, -InverseErf[z]^2]^Subscript[j, i], {i, 2, n}], {Subscript[j, n], 0, n}], {Subscript[j, 2], 0, n}] /; Element[n, Integers] && n >= 0
No reference, outside of Wolfram functions, for the formula was found, but the top FAQ question implies that the formula has been verified with Mathematica. Therefore, the analytic inverse function’s expansion about $z=\erf(1)=0.8427…$ should be:
$$\erf^{-1}(z)\mathop=^?\sum_{n=0}^\infty\frac{(z-\erf(1))^n}{n!}\left(\delta_n+\left(\frac{\sqrt\pi}2e\right)^n\sum_{j_2=0}^n\cdots\sum_{j_n=0}^n (-1)^{\sum\limits_{k=2}^\infty j_k}\delta_{\sum\limits_{k=2}^n(k-1)j_k,n-1}\left(n-1+\sum_{k=2}^n j_k\right)!\prod_{k=2}^n \frac1{j_k!}\left(\frac{e\sqrt\pi2^{k-1}}{k!}\right)^{j_k}\,_2\tilde{\text F}_2\left(\frac12,1;1-\frac k2,\frac{3-k}2;-1\right)^{j_k}\right)= 1+\frac{e\sqrt\pi}2(z-\erf(1))+\sum_{n=2}^\infty\frac{(z-\erf(1))^n}{n!}\left(\delta_n+\left(\frac{\sqrt\pi}2e\right)^n\sum_{j_2=0}^n\cdots\sum_{j_n=0}^n (-1)^{\sum\limits_{k=2}^nj_k}\delta_{\sum\limits_{k=2}^n(k-1)j_k,n-1}\left(n-1+\sum_{k=2}^n j_k\right)!\prod_{k=2}^n \frac1{j_k!}\left(\frac{e\sqrt\pi2^{k-1}}{k!}\right)^{j_k}\,_2\tilde{\text F}_2\left(\frac12,1;1-\frac k2,\frac{3-k}2;-1\right)^{j_k}\right) $$
There is a Mathematica formula download for $\erf^{-1}(z)$ to test the series expansion, but I do not have Mathematica. Also, the series about $z\to0$ may diverge since $\lim\limits_{z\to0}\frac{e^{\erf^{-1}(z)^2}}{\erf^{-1}(z)^{k-1}}=\infty,k>1$ and the $\sum\limits_{j_2=0}^n\cdots\sum\limits_{j_n=0}^n (-1)^{\sum\limits\limits_{k=2}^n j_k}\delta_{\sum\limits\limits_{k=2}^n(k-1)j_k,n-1}\left(n-1+\sum\limits_{k=2}^n j_k\right)! $ reminds one of the explicit series reversion formula in formula $(11)$. If this series works, then we have an explicit series representation of the inverse error function and possibly an analytic continuation of the function.
Does this $n$th derivative formula give the Taylor series coefficients $a_n$ for $\erf^{-1}(z)=\sum\limits_{n=0}^\infty a_n\frac{(z-z_0)^n}{n!}$? If so, how does one derive an explicit series about $z=0$?
This final question is optional, but @Steven Clark suggested the radius of convergence of $\erf^{-1}(z)=\sum\limits_{n=0}^\infty\frac{(z-z_0)^n}{n!}\frac{d^n}{dz^n}\erf^{-1}(z)\big|_{z=z_0}\text{ using }(1)$ is needed to know convergence and where an analytic continuation is possible.
This is not an answer.
Since we faced the problem of the inverse of the error function quite many years ago in my former research group, I give you the results for the inversion of $$z=\text{erf}(x) \qquad \text{around} \qquad x=a$$
It write $$\color{red}{x=\sum_{n=0}^\infty \frac{\pi ^{\frac n2}\, e^{a^2 n} }{n!\,2^{\left\lfloor \frac{n+1}{2}\right\rfloor } }\,P_n(a)\,\big[z-\text{erf}(a)\big]^n}$$ the first polynomials being $$\left( \begin{array}{cc} n & P_n(a) \\ 0 & a \\ 1 & 1 \\ 2 & a \\ 3 & 4 a^2+1 \\ 4 & a \left(12 a^2+7\right) \\ 5 & 96 a^4+92 a^2+7 \\ 6 & a \left(480 a^4+652 a^2+127\right) \\ 7 & 5760 a^6+10224 a^4+3480 a^2+127 \\ 8 & a \left(40320 a^6+88848 a^4+44808 a^2+4369\right) \\ 9 & 645120 a^8+1703808 a^6+1161168 a^4+204328 a^2+4369 \\ 10 & a \left(5806080 a^8+17914752 a^6+15561936 a^4+4161288 a^2+243649\right) \end{array} \right)$$ which show interesting patterns. After reading you post, it is almost obvious that these polynomials correspond to some hypergeometric functions (which I did not identify)
For $a=1$, the $P_n(1)$ generate the sequence $$\{1,1,1,5,19,195,1259,19591,178345,3718793,43687705,1141949277,16433602763\}$$ not found in $OEIS$.