the standard error of a standard deviation is given as : s/n^(1/2). Would the standard error of kurtosis and skewness follow the same idea? For example, se of kurtosis = kurtosis/n^(1/2)?
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The answer is no. Your "standard error of a standard deviation" is also incorrect (that's the standard error of the sample mean). You can calculate the distributions (and hence vasiance/standard errors) of higher sample moments assuming certain distributions, for example here for the Normal distribution--notice that it's pretty complicated. If you want to estimate the standard error of the sample kurtosis my recommendation would be to do bootstrapping.
Edit: Since you also mentioned skewness, you should note in particular that skewness can be negative, so the standard error according to your formula would end up being negative. That's not cool.