Does the sum $\frac{1}{3} + \frac{1}{3^{1+1/2}}+\cdots$ have a closed form?

If this was something you just came up with, it is highly unlikely there is any obtainable closed form expression. Checking the number Wolfram|Alpha generates from sum (1/(3^(sum (1/k) from k=1 to n))) from n=1 to infinity in an inverse symbolic calculator, I did not find anything.

We can certainly impose some bounds on the value of the sum, via the asymptotic expansion $$H_n \sim \log n + \gamma + \frac{1}{2n} - \frac{1}{12n^2} + \cdots. \tag{1}$$ The crudest bound is to note for $0 < z < 1$ the sum $$f(z) = \sum_{n=1}^\infty z^{H_n}$$ is dominated by $$\begin{align*} f(z) &< z^\gamma \sum_{n=1}^\infty z^{\log n} \\ &= z^\gamma \sum_{n=1}^\infty e^{\log z \log n}\\ &= z^\gamma \sum_{n=1}^\infty n^{\log z} \\ &= z^\gamma \zeta(-\log z). \tag{2} \end{align*}$$ For $z = 1/3$, this gives us the comparison $$f(1/3) \approx 5.34863 < 5.688508.$$ More terms of the asymptotic expansion $(1)$ can be used to speed the computation. However, we must be careful since $(1)$ is centered around $n = \infty$, so convergence is poor for small $n$; we can compensate by computing the initial terms precisely, then using the asymptotic expanison for large $n$, resulting in rapid convergence.