This can be done through Bayesian inference, on a maximum likelihood estimation basis. Assuming a (log)normal distribution, you could calculate the likelihood that any prior (however stupid) estimate of mu and sigma (and thus variance). Since the normal distribution is its own conjugate prior, one can always come up with an iterative better guess for the parameters. But it's iterative; and the update formulae ain't pretty.
https://www.statlect.com/fundamentals-of-statistics/normal-distribution-Bayesian-estimation
A simpler way - that is unbiased - would be fall back on the observation that the variance calculation can be decomposed into a sum of the squares versus mean formulation. Both of which are easy to meausre.
Your mean is easy to compute, give the aggregation problem. Just third/log-third the quarterly one. Then calculate the sum of the squares of your returns - irrespective of timeframe. This just assumes temporall return independence, which is implicit measuring monthly returns in the first place. If this, then quarterly returns should be root-3 more volatile than monthly returns; and monthly variances 3x. The sum-of-squares follows.
https://www.sciencebuddies.org/science-fair-projects/science-fair/variance-and-standard-deviation#:~:text=The%20variance%20(%CF%832)%2C,in%20the%20distribution%20(N).&text=You%20take%20the%20sum%20of,in%20the%20distribution%20(N).
Given this average and sum of squares, the variance is as below: