I think all you need is to get exponent of the arithmetic mean of log-log of your series.
The geometric means is basically arithmetic mean on log series: $$\bar y=\frac 1 n\sum_{i=1}^n y_i$$ $$\frac 1 n\sum_{i=1}^n\log x_i=\frac 1 n\log\prod_{i=1}^nx_i= \log\left(\prod_{i=1}^nx_i\right)^{\frac 1 n}$$ $$\exp\left(\frac 1 n\sum_{i=1}^n\log x_i\right)\equiv e^{\bar x}= \left(\prod_{i=1}^nx_i\right)^{\frac 1 n}$$
Using the same logic idea, you can try the arithmetic mean of log-log series: $$\frac 1 n\sum_{i=1}^n\log\log x_i=\frac 1 n\log\prod_{i=1}^n\log x_i= \log\left(\prod_{i=1}^n\log x_i\right)^{\frac 1 n}$$
The last term is linked to something called a "nested exponent": $$\prod_{i=1}^n\log x_i=\log x_1^{{x_2}^{{\dots}^{x_n}}}$$