I would say that since sample mean is $\overline{X}=\frac{\sum{X_i}}{n}$, its convergence rate is 1/n, which is also the convergence rate of sample variance. But the convergence rate of sample standard deviation $ 1/\sqrt{n} $.
I am somewhat pessimistic about a such non-parametric method, at least without the introduction of some sort of constraints on the underlying distribution.
My reasoning for this is that there will always be a distribution that breaks the true coverage probability for any finite $n$ (although as $n \rightarrow \infty$, this distribution will become more and more pathological), or the confidence interval will have to be arbitrarily large.
To illustrate, you could imagine a distribution that looks like a normal up to some value $\alpha$, but after $\alpha$ becomes extremely right skewed. This can have unbounded influence on the distribution's mean and as you push $\alpha$ out as far as possible, this can have arbitrarily small probability of making it into your sample. So you can imagine that for any $n$, you could pick an $\alpha$ to be so large that all points in your sample have extremely high probability of looking like it comes from a normal distribution with mean = 0, sd = 1, but you can also have any true mean.
So if you're looking for proper asymptotic coverage, of course this can be achieved by the CLT. However, your question implies that you are (quite reasonably) interested in the finite coverage. As my example shows, there will always be a pathological case that ruins any finite length CI.
Now, you still could have a non-parametric CI that achieves good finite coverage by adding constraints to your distribution. For example, the log-concave constraint is a non-parametric constraint. However, it seems inadequate for your problem, as log-normal is not log-concave.
Perhaps to help illustrate how difficult your problem could be, I've done unpublished work on a different constraint: inverse convex (if you click on my profile, I have a link to a personal page that has a preprint). This constraint includes most, but not all log-normals. You can also see that for this constraint, the tails can be "arbitrarily heavy", i.e. for any inverse convex distribution up to some $\alpha$, you can have heavy enough tails that the mean will be as large as you like.
Best Answer
The mean is an ordinary least squares estimator, and so it's BLUE - it's the best linear unbiased estimator. Best, in this case means that it has the smallest sampling variance of any unbiased estimator (including the median).
The proof is on the Wikipedia page for Gauss-Markov theorem: http://en.wikipedia.org/wiki/Gauss%E2%80%93Markov_theorem
Another way to think about it is that the mean uses all of the information available in the sample, the median is just one value.
Another way to look at it is to run a simulation, calculating the median and mean for a large number of samples, and to look at their standard deviation:
The mean has a smaller variance (sd), which means it has narrower confidence intervals.