Does standard deviation and its confidence interval consider the stochastic variability of data? If we compute the standard deviation of a data set composed of a single feature and then compute its confidence interval, then can we say that these computations have considered the stochastic variability of the data? What exactly is stochastic variability of the data?
 A: I presume the intent of 'stochastic variability' is simply 'random variation', and by "its confidence interval" you mean a confidence interval for $\sigma$ (since that's the only parameter you mention and you don't form a confidence interval for data).
If I understood the intent of the question correctly, then the standard deviation not only considers the variability, it's a measure of "how big" it is. 
A suitable confidence interval for the standard deviation would be based in some direct sense on how variable that data is.
For example, if we assume normal random variables with constant mean and variance and mutual independence, then an interval for the standard deviation could be obtained by taking the square roots of the usual limits of the interval for the variance:
$\left[\frac{(n-1)s^2}{Q(1-\alpha/2)},\frac{(n-1)s^2}{Q(\alpha/2)}\right]$
where $Q(p)$ is the $p$th quantile of a $\chi^2_{(n-1)}$
As you see, the variance-interval itself depends on the sample variance.
Similarly a $1-\alpha$ interval for $\sigma$ would be (after taking out the common factor of $s$:
$\qquad s\left[\sqrt{\frac{(n-1)}{Q(1-\alpha/2)}},\sqrt{\frac{(n-1)}{Q(\alpha/2)}}\right]$
which clearly depends on $s$, which as mentioned before is a measure of the amount of random variability in the sample and an estimate of the amount of random variability in the population.
