Use of standard error of the mean when the components are not identically-distributed Let {$X_1$, ..., $X_n$} be a random sample of size n in which the different elements are measurements drawn from $m$ different populations with different distributions. From the Central Limit Theorem, as long as $n \gg m$ and $n  \to \infty$, the values of $\bar{X}$ obtained by repeating this experiment will follow a normal distribution irrespective of the properties of the different $m$ starting distributions.
Does this mean that it is valid to use the standard error of the mean as representative of the uncertainty on the $\bar{X}$? The different $\bar{X}$ are not independent measurements of the same variable, but the fact they they follow a normal distribution would make me believe it might be legitimate to do so.
 A: All we need are basic properties of expectation and variance.
Let $X_i\sim F_i$ be independent with mean and variance $\mu_i$ and $\sigma^2_i$ respectively.
Then $E(\bar X)=\bar \mu = \frac{_1}{^n} \sum_{i=1}^n \mu_i$ (linearity of expectation)
and $\text{Var}(\bar X) = \text{Var}(\frac{_1}{^n} \sum_{i=1}^nX_i) =\frac{_1}{^{n^2}} \sum_{i=1}^n \text{Var}(X_i)=\frac{_1}{^{n^2}} \sum_{i=1}^n \sigma^2_i$.
From that, you can immediately compute the standard error of the mean as $\frac{1}{^n}\sqrt{\sum_{i=1}^n \sigma^2_i}$.
(Indeed, you don't have to have independence, uncorrelated variables will do.)
If the $X_i$ are correlated, the calculation can still be done (as long as you also know the pairwise covariances or correlations), it's just a little more complicated - 
specifically, $\text{Var}(\sum X_i) = \sum \text{Var}(X_i) + 2\sum\sum_{i< j} \text{Cov}(X_i,X_j)$ -- which is again straight from basic properties of variance.
These are not asymptotic results, as the CLT is, but exact, finite sample results that hold whether you have normal distributions or not. For example, if $X_1$ is $\text{Exponential}(1)$ and $X_2$ is $\text{Uniform}(0,1)$, independent of $X_1$, their average is not even close to normal, but the standard error of their average is still given by the formula I gave above.
Here's a  histogram of simulated values and theoretical density for the average of a standard exponential and standard uniform:

$E(X_1)=1$ and $E(X_2)=\frac{1}{2}$. $E(\frac{X_1+X_2}{2})  = \frac{1+\frac{1}{2}}{2} = \frac{3}{4}$ (simulated = 0.74975).
$\text{Var}(X_1)=1$ and $\text{Var}(X_2)=\frac{1}{12}$. $\text{Var}(\frac{X_1+X_2}{2})  = \frac{1}{4} (1+\frac{1}{12}) = \frac{13}{48}=0.27083...$ (simulated = 0.273).
Standard error of the mean is then $\sqrt\frac{13}{48}\approx 0.5204$.


Does this mean it is valid to use the standard error of the mean as representative of the uncertainty on the $\bar X$

That depends on what you regard as 'representative of the uncertainty'. There may be situations when you don't regard the standard error of $\bar X$ as the most appropriate way to measure the uncertainty of $\bar X$. If you want the standard error, you have it.
