I am looking for a simplified form of the infinite sum of harmonic exponentials
$$ f(x) = \sum_{n=0}^{\infty}x^{H_n} = 1 + x + x^{3/2} + x^{11/6} + x^{25/12} + x^{137/60} + x^{49/20} + \ldots\\ =1+x\left( 1+x^{1/2}\left( 1+x^{1/3}\left( 1+x^{1/4}\left( 1+x^{1/5}\left( 1+x^{1/6}\left( 1+\ldots \right)\right)\right)\right)\right)\right) $$ for $x\in[0,1/e)$.
Can it be reduced to a standard special function?
Update: a stable way of calculating $f(x)$ numerically
Inspired by Claude Leibovici's answer, we write $$ f(x) = 1 + \sum_{n=1}^{\infty}x^{\gamma+\log(n)} - \underbrace{\sum_{n=1}^{\infty}\left[x^{\gamma+\log(n)}-x^{H_n}\right]}_{g(x)}\\ = 1+x^{\gamma}\zeta[-\log(x)]-g(x) $$ with $g(x)$ converging rapidly on the entire interval $x\in[0,1/e]$ (and beyond) without singularities:
Specifically, $$ g(1/e) = \sum_{n=1}^{\infty}\left[\frac{e^{-\gamma}}{n}-e^{-H_n}\right] \approx 0.35370447459102146589\ldots\\ g'(1/e) \approx 0.38530115152397563963\ldots\\ g''(1/e) \approx -0.82751608018475931083\ldots $$ give the approximate series expansion of $f(x)$ around $x=1/e$, $$ f(x) \approx \frac{e^{-\gamma-1}}{1/e-x} + 0.36556578362553594918\ldots + (1/e-x)\cdot0.11500038261356357678\ldots + (1/e-x)^2\cdot0.11786065053337882776\ldots +O[(1/e-x)^3] $$
Unfortunately, neither WolframAlpha nor AskConstants recognize any of these numbers.
As @Raffaele commented, if you use the approximation $$H_n\sim \log( n)+\gamma$$
Using what you wrote (notice that $H_7=2.59$ while $\log(7)+\gamma=2.52$) $$f(x)\sim1 + x + x^{3/2} + x^{11/6} + x^{25/12} + x^{137/60} + x^{49/20} + x^\gamma\sum_{n=7}^{\infty}n^{\log(x)}$$ $$\sum_{n=7}^{\infty}n^{\log(x)}=\zeta (-\log (x))-\sum_{k=1}^6 k^{\log(x)}$$ which converges for $x\lt \frac 1e$.
Let us compare for a few values of $x$ $$\left( \begin{array}{ccc} x & \text{approximation} & \text{exact} \\ 0.00 & 1.00000 & 1.00000 \\ 0.02 & 1.02424 & 1.02421 \\ 0.04 & 1.05408 & 1.05391 \\ 0.06 & 1.08957 & 1.08911 \\ 0.08 & 1.13148 & 1.13055 \\ 0.10 & 1.18100 & 1.17937 \\ 0.12 & 1.23976 & 1.23718 \\ 0.14 & 1.30998 & 1.30621 \\ 0.16 & 1.39482 & 1.38955 \\ 0.18 & 1.49874 & 1.49168 \\ 0.20 & 1.62838 & 1.61920 \\ 0.22 & 1.79396 & 1.78233 \\ 0.24 & 2.01210 & 1.99766 \\ 0.26 & 2.31175 & 2.29413 \\ 0.28 & 2.74822 & 2.72704 \\ 0.30 & 3.44199 & 3.41684 \\ 0.32 & 4.71483 & 4.68530 \\ 0.34 & 7.81187 & 7.77753 \\ 0.36 & 26.6198 & 26.5802 \end{array} \right)$$ which seems to be quite decent.
So, you can write $$\sum_{n=0}^{\infty}x^{H_n}=\sum_{n=0}^{m}x^{H_n}+x^\gamma\Big[\zeta (-\log (x))-\sum_{k=1}^{m+1} k^{\log(x)}\Big]\tag 1$$
As @Roman commented, if you take a large value of $m$, the asymptotics is $$f(x)\sim \sum_{n=0}^{m}x^{H_n}-\frac{m\, x^{\gamma+\log (m)}}{1+\log(x)}$$
Edit
Equation $(1)$ can be simplified ; in terms of the Hurwitz zeta function, it is $$f(x)=\sum_{n=0}^{\infty}x^{H_n}=\sum_{n=0}^{m}x^{H_n}+x^\gamma \,\zeta (-\log (x),m+1)$$ and expanded as series $$f(x)=\sum_{n=0}^{m}x^{H_n}-x^{\gamma+\log(m)} g(t)\qquad \text{where} \qquad t=\log(x)$$ $$g(t)=\frac{m}{t+1}+\frac{1}{2}+\frac{t}{12 m}-\frac{(t-2) (t-1) t}{720 m^3}+\frac{(t-4) (t-3) (t-2) (t-1) t}{30240 m^5}-\frac{(t-6) (t-5) (t-4) (t-3) (t-2) (t-1) t}{1209600 m^7}+O\left(\frac{1}{m^9}\right)$$