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If you work in terms of indicator variables (i.e. $Z_i = 1$ if $X_i \leq x$ and $0$ otherwise), you can directly apply the Central limit theorem to a mean of $Z$'s, and by using the Delta method, turn that into an asymptotic normal distribution for $F_X^{-1}(\bar{Z})$, which in turn means that you get asymptotic normality for fixed quantiles of $X$.

So not just the median, but quartiles, 90th percentiles, ... etc.

Loosely, if we're talking about the $q$th sample quantile in sufficiently large samples, we get that it will approximately have a normal distribution with mean the $q$th population quantile $x_q$ and variance $q(1-q)/(nf_X(x_q)^2)$.

Hence for the median ($q = 1/2$), the variance in sufficiently large samples will be approximately $1/(4nf_X(\tilde{\mu})^2)$.

You need all the conditions along the way to hold, of course, so it doesn't work in all situations, but for continuous distributions where the density at the population quantile is positive and differentiable, etc, ...

Further, it doesn't hold for extreme quantiles, because the CLT doesn't kick in there (the average of Z's won't be asymptotically normal). You need different theory for extreme values.

If you work in terms of indicator variables (i.e. $Z_i = 1$ if $X_i \leq x$ and $0$ otherwise), you can directly apply the Central limit theorem to a mean of $Z$'s, and by using the Delta method, turn that into an asymptotic normal distribution for $F_X^{-1}(\bar{Z})$, which in turn means that you get asymptotic normality for fixed quantiles of $X$.

So not just the median, but quartiles, 90th percentiles, ... etc.

Loosely, if we're talking about the $q$th sample quantile in sufficiently large samples, we get that it will approximately have a normal distribution with mean the $q$th population quantile $x_q$ and variance $q(1-q)/(nf_X(x_q)^2)$.

Hence for the median ($q = 1/2$), the variance in sufficiently large samples will be approximately $1/(4nf_X(\tilde{\mu})^2)$.

You need all the conditions along the way to hold, of course, so it doesn't work in all situations, but for continuous distributions where the density at the population quantile is positive and differentiable, etc, ...

Further, it doesn't hold for extreme quantiles, because the CLT doesn't kick in there (the average of Z's won't be asymptotically normal). You need different theory for extreme values.

Edit: whuber's critique is correct; this would work if $x$ were a population median rather than a sample median. The argument needs to be modified to actually work properly.

If you work in terms of indicator variables (i.e. $Z_i = 1$ if $X_i \leq x$ and $0$ otherwise), you can directly apply the Central limit theorem to a mean of $Z$'s, and by using the Delta method, turn that into an asymptotic normal distribution for $F_X^{-1}(\bar{Z})$, which in turn means that you get asymptotic normality for fixed quantiles of $X$.

So not just the median, but quartiles, 90th percentiles, ... etc.

Loosely, if we're talking about the $q$th sample quantile in sufficiently large samples, we get that it will approximately have a normal distribution with mean the $q$th population quantile $x_q$ and variance $q(1-q)/(nf_X(x_q)^2)$.

Hence for the median ($q = 1/2$), the variance is $1/(4nf_X(\tilde{\mu})^2)$.

You need all the conditions along the way to hold, of course, so it doesn't work in all situations, but for continuous distributions where the density at the population quantile is positive and differentiable, etc, ...

Further, it doesn't hold for extreme quantiles, because the CLT doesn't kick in there (the average of Z's won't be asymptotically normal). You need different theory for extreme values.

If you work in terms of indicator variables (i.e. $Z_i = 1$ if $X_i \leq x$ and $0$ otherwise), you can directly apply the Central limit theorem to a mean of $Z$'s, and by using the Delta method, turn that into an asymptotic normal distribution for $F_X^{-1}(\bar{Z})$, which in turn means that you get asymptotic normality for fixed quantiles of $X$.

So not just the median, but quartiles, 90th percentiles, ... etc.

Loosely, if we're talking about the $q$th sample quantile in sufficiently large samples, we get that it will approximately have a normal distribution with mean the $q$th population quantile $x_q$ and variance $q(1-q)/(nf_X(x_q)^2)$.

Hence for the median ($q = 1/2$), the variance in sufficiently large samples will be approximately $1/(4nf_X(\tilde{\mu})^2)$.

You need all the conditions along the way to hold, of course, so it doesn't work in all situations, but for continuous distributions where the density at the population quantile is positive and differentiable, etc, ...

Further, it doesn't hold for extreme quantiles, because the CLT doesn't kick in there (the average of Z's won't be asymptotically normal). You need different theory for extreme values.

If you work in terms of indicator variables (i.e. $Z_i = 1$ if $X_i \leq x$ and $0$ otherwise), you can directly apply the Central limit theorem to a mean of $Z$'s, and by using the Delta method, turn that into an asymptotic normal distribution for $F_X^{-1}(\bar{Z})$, which in turn means that you get asymptotic normality for fixed quantiles.

So not just the median, but quartiles, 90th percentiles, ... etc.

Loosely, if we're talking about the $q$th sample quantile in sufficiently large samples, we get that it will approximately have a normal distribution with mean the $q$th population quantile $x_q$ and variance $q(1-q)/(nf_X(x_q)^2)$.

Hence for the median ($q = 1/2$), the variance is $1/(4nf_X(\tilde{\mu})^2)$.

You need all the conditions along the way to hold, of course, so it doesn't work in all situations, but for continuous distributions where the density at the population quantile is positive and differentiable, etc, ...

Further, it doesn't hold for extreme quantiles, because the CLT doesn't kick in there (the average of Z's won't be asymptotically normal). You need different theory for extreme values.

If you work in terms of indicator variables (i.e. $Z_i = 1$ if $X_i \leq x$ and $0$ otherwise), you can directly apply the Central limit theorem to a mean of $Z$'s, and by using the Delta method, turn that into an asymptotic normal distribution for $F_X^{-1}(\bar{Z})$, which in turn means that you get asymptotic normality for fixed quantiles of $X$.

So not just the median, but quartiles, 90th percentiles, ... etc.

Loosely, if we're talking about the $q$th sample quantile in sufficiently large samples, we get that it will approximately have a normal distribution with mean the $q$th population quantile $x_q$ and variance $q(1-q)/(nf_X(x_q)^2)$.

Hence for the median ($q = 1/2$), the variance is $1/(4nf_X(\tilde{\mu})^2)$.

You need all the conditions along the way to hold, of course, so it doesn't work in all situations, but for continuous distributions where the density at the population quantile is positive and differentiable, etc, ...

Further, it doesn't hold for extreme quantiles, because the CLT doesn't kick in there (the average of Z's won't be asymptotically normal). You need different theory for extreme values.