MAD equivalent for standard error As far as I know, one can calculate the relative standard error from the standard deviation of a data sample. I am looking for the Median Absolute Deviation equivalent for standard error. 
Does one exist? Also which methods exist to calculate the relative standard error, that do not require the standard deviation?
 A: When you say standard error, you should be talking about the standard error of something, such as the standard error of the sample mean. 
Similarly you could for example talk about the median absolute deviation of the sample median. It is possible to calculate this, at least as an approximation for large samples.  
It is well known that for a continuous random variable with population median $m$, continuous probability density of the median $f(m)$ and a large odd sample size $n$, the sample median is approximately normally distributed with median $m$ and variance $\frac{1}{4 n f(m)^2}$, i.e. with median absolute deviation approximately $\dfrac{\Phi^{-1}\left(\frac34 \right)}{2  \sqrt{n} f(m)}$ where $\frac{\Phi^{-1}\left(\frac34 \right)}{2} \approx 0.337$.
If you want to have this as a relative median absolute deviation of the sample median, then presumably you divide by $m$.
A: An alternative approach is to look at the quantiles of the the sample: order the observations so $x_1 \le x_2 \le x_3 \le \cdots \le x_n$ and (for reasonably large samples from a continuous distribution) there is a 95% probability that the interval $$\left[x_{\frac{n}{2} - 0.98 \sqrt{n}}, x_{\frac{n}{2} + 0.98 \sqrt{n}}\right]$$ contains the population median.
Clearly ${\frac{n}{2} \pm 0.98 \sqrt{n}}$ may not be an integer: you can round outwards to be conservative or use one of the many possibilities for interpolating quantiles: you are now looking for $Q_p$ where $p =  {\frac{1}{2} \pm 0.98 /\sqrt{n}}$.
Again for relative, you can divide by the median. 
