If the distribution is symmetric, then the mean is equal to the median. So then you can meta-analyze the means, since that also gives you an estimate of the median. In addition, it is more efficient.
It is also possible to meta-analyze medians directly. The large-sample variance of a sample median ($m$) from a normal distribution is $$\mbox{Var}[m] = \frac{\pi \sigma^2}{2n}.$$ An estimate of the sampling variance can be obtained by replacing $\sigma^2$ with the observed sample variance. So, given multiple medians and corresponding variances thereof, one can easily proceed with a meta-analysis of these values. But again, it would be more efficient to meta-analyze the means then.
If the raw data did not come from a normal distribution (and especially if the distribution is not symmetric), then things are different. Then meta-analyzing means and medians are really different things. However, if the data are not normally distributed, but follow some other distribution with density function $f(x)$, then the variance equation above is not correct. The more general equation for the large-sample variance is $$\mbox{Var}[m] = \frac{1}{4nf(M)^2},$$ so $f(M)$ is the density at the true median (see, for example, https://en.wikipedia.org/wiki/Median#Sampling_distribution or https://stats.stackexchange.com/a/45143/1934). Note that the density of the normal distribution at the median (= mean) is $1/\sqrt{2 \sigma^2 \pi}$, so just plug that into the equation and you get the variance equation given earlier.
So if you want to meta-analyze medians, you first need to have some kind of idea what density function would be applicable for a given dataset. Then you can compute (or rather: estimate) the sampling variance and then again proceed with a standard meta-analysis.