I think all you need is to get the arithmetic mean of log-log of your series. Here's why.
The geometric mean is basically an exponent of arithmetic mean of log series: $$\bar y=\frac 1 n\sum_{i=1}^n y_i$$ $$\frac 1 n\sum_{i=1}^n\log x_i=\frac 1 n\log\prod_{i=1}^nx_i= \log\left(\prod_{i=1}^nx_i\right)^{\frac 1 n}$$ $$\exp\left(\frac 1 n\sum_{i=1}^n\log x_i\right)\equiv e^{\bar x}= \left(\prod_{i=1}^nx_i\right)^{\frac 1 n}$$
If you have the base normal distribution $\mathcal N(\mu,\sigma^2)$, which forms a lognormal distribution then the geometric mean of the lognormal series is going to give you a decent estimator $\hat\mu$.
Using the same logic, you can try the arithmetic mean of log-log series: $$\hat\mu'=\frac 1 n\sum_{i=1}^n\log\log x_i$$
This will give a decent estimator $\hat\mu'$ of the mean of the base normal of your log-lognormal distribution. I won't say that it's the most optimal estimator, not sure about that, but it will be a sensible one.