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Perhaps to help illustrate how difficult your problem could be, I've done unpublished work on a different constraint: inverse convex (if you click on my profile, I have a link to a personal page that has a preprint).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 question is related to the fundamental construction of confidence intervals, and when it comes to bootstrapping, the answer depends upon which bootstrapping method that is used.
Consider the following setup: $\hat{\theta}$ is an estimator of a real valued parameter $\theta$ with (an estimated) standard deviation $\text{se}$, then a standard 95% confidence interval based on a normal $N(\theta, \text{se}^2)$ approximation is $$\hat{\theta} \pm 1.96 \text{se}.$$ This confidence interval is derived as the set of $\theta$'s that fulfill $$z_{1} \leq \hat{\theta} - \theta \leq z_2$$ where $z_1 = -1.96\text{se}$ is the 2.5% quantile and $z_2 = 1.96\text{se}$ is the 97.5% quantile for the $N(0, \text{se}^2)$-distribution. The interesting observation is that when rearranging the inequalities we get the confidence interval expressed as $$\{\theta \mid \hat{\theta} - z_2 \leq \theta \leq \hat{\theta} - z_1 \} = [\hat{\theta} - z_2, \hat{\theta} - z_1].$$ That is, it is the lower 2.5% quantile that determines the right end point and the upper 97.5% quantile that determines the left end point.
If the sampling distribution of $\hat{\theta}$ is right skewed compared to the normal approximation, what is then the appropriate action? If right-skewed means that the 97.5% quantile for the sampling distribution is $z_2 > 1.96\text{se}$, the appropriate action is to move the left end point further to the left. That is, if we stick to the standard construction above. A standard usage of the bootstrap is to estimate the sampling quantiles and then use them instead of $\pm 1.96 \text{se}$ in the construction above.
However, another standard construction used in bootstrapping is the percentile interval, which is $$[\hat{\theta} + z_1, \hat{\theta} + z_2].$$ in the terminology above. It is simply the interval from the 2.5% quantile to the 97.5% quantile for the sampling distribution of $\hat{\theta}.$ A right-skewed sampling distribution of $\hat{\theta}$ implies a right-skewed confidence interval. For the reasons mentioned above, this appears to me to be a counter-intuitive behavior of percentile intervals. But they have other virtues, and are, for instance, invariant under monotone parameter transformations.
The BCa (bias-corrected and accelerated) bootstrap intervals as introduced by Efron, see e.g. the paper Bootstrap Confidence Intervals, improve upon the properties of percentile intervals. I can only guess (and google) the quote the OP post, but maybe BCa is the appropriate context. Citing Diciccio and Efron from the paper mentioned, page 193,
where (2.3) is the definition of the BCa intervals. The quote posted by the OP may refer to the fact that BCa can shift confidence intervals with a right-skewed sampling distribution further to the right. It is difficult to tell if this is the "correct action" in a general sense, but according to Diciccio and Efron it is correct in the setup above in the sense of producing confidence intervals with the correct coverage. The existence of the monotone transformation $m$ is a little tricky, though.