# Is there an analogue for standard error of the mean based on mean absolute deviation?

We can estimate the standard error (SE) of the sample mean as the sample standard deviation divided by the square root of the number of samples, cf. https://en.wikipedia.org/wiki/Standard_error#Standard_error_of_the_sample_mean which writes it as:

$${\sigma }_{\bar {\mathbf{x}}}\ \approx {\frac {\sigma _{\mathbf{x}}}{\sqrt {N}}}}$$

Does there exist a similar measure (where the number of samples is involved) to estimate the standard error of the mean based on mean absolute deviation (MAD)? By MAD I mean the mean of the absolute value of the deviation from the mean:

$$\displaystyle\mathrm{MAD}(\mathbf{x}) = \frac{1}{N}\sum_i^{N} \mathrm{Abs}(x_i-\bar{\mathbf{x}})$$

The MAD is, in my mind, a similar quantity to the SD (in that it measures some sort of spread), but is not equal to the SD in the limit $$N\to\infty$$. Would it be terribly wrong to write the MAD-based SE as something like $$\displaystyle\frac{\mathrm{MAD}(\mathbf{x})}{\sqrt{N}}$$?