Suppose I'm running an experiment that can have 2 outcomes, and I'm assuming that the underlying "true" distribution of the 2 outcomes is a binomial distribution with parameters $n$ and $p$: ${\rm Binomial}(n, p)$. 

I can compute the standard error, $SE_X = \frac{\sigma_X}{\sqrt{n}}$,  from the form of the variance of  ${\rm Binomial}(n, p)$: $$ \sigma^{2}_{X} = npq$$ 
where $q = 1-p$. So, $\sigma_X=\sqrt{npq}$. For the standard error I get: $SE_X=\sqrt{pq}$, but I've seen somewhere that $SE_X = \sqrt{\frac{pq}{n}}$. What did I do wrong?

It seems like you're using $n$ twice in two different ways - both as the sample size and as the number of bernoulli trials that comprise the Binomial random variable; to eliminate any ambiguity, I'm going to use $k$ to refer to the latter. 

If you have $n$ independent samples from a ${\rm Binomial}(k,p)$ distribution, the variance of their sample mean is

$$ {\rm var} \left( \frac{1}{n} \sum_{i=1}^{n} X_{i} \right) = \frac{1}{n^2} \sum_{i=1}^{n} {\rm var}( X_{i} ) = \frac{ n {\rm var}(X_{i}) }{ n^2 } = \frac{ {\rm var}(X_{i})}{n} = \frac{ k pq }{n} $$

where $q=1-p$ and $\overline{X}$ is the same mean. This follows since 

(1) ${\rm var}(cX) = c^2 {\rm var}(X)$, for any random variable, $X$, and any constant $c$. 

(2) the variance of a sum of independent random variables equals the sum of the variances. 

The standard error of $\overline{X}$is the square root of the variance: $\sqrt{\frac{ k pq }{n}}$. Therefore, 


When $k = n$, you get the formula you pointed out: $\sqrt{pq}$
When $k = 1$, and the Binomial variables are just bernoulli trials, you get the formula you've seen elsewhere: $\sqrt{\frac{pq }{n}}$

