Suppose you want to estimate the mean $\mu$ of a normal population using the
mean $\bar X$ of a random sample $X_1, X_2, \dots, X_n$ of size $n$
from the population.
The term 'standard error' usually refers to the the standard deviation
of an estimator. In the current situation the standard error of $\bar X$
is $\sigma/\sqrt{n},$ where $\sigma$ is the standard deviation of the population.
However, if $\sigma$ is unknown and estimated by the sample standard deviation
$S= \sqrt{\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2},$ then
$T = \frac{\bar X = \mu}{S/\sqrt{n}}$ has Student's t distribution with
$\nu = n-1$ degrees of freedom, $\mathsf{T}(\nu).$
This fact is used to make a 95% confidence for (estimating) $\mu$ of the form
$\bar X \pm t^*\frac{S}{\sqrt{n}},$ where $t^*$ cuts probability $0.025= 2.5\%$ from
the upper tail of the distribution $\mathsf{T}(\nu).$ [By the symmetry of the t distribution $-t^*$ cuts 2.5% of the probability from the lower tail of the distribution, leaving 95% in the middle.]
In this situation
the estimated standard deviation of $\bar X$ is $S/\sqrt{n}.$ Usually
the word estimated is left out because as soon as you see $S$ you know
it's an estimate of $\sigma.$
You might say you have two standard errors
of the sample mean $\bar X,$ the theoretical one $\sigma/\sqrt{n},$ and the
estimated one, $S/\sqrt{n}.$ However, if $\sigma$ is unknown, only the (estimated)
standard error $S/\sqrt{n}$ is of practical use---in making a confidence
interval as above or in testing a hypothesis about $\mu.$
Note: There are situations in which $\mu$ is unknown and $\sigma$ is known.
Then a 95% confidence interval for $\mu$ is $\bar X \pm 1.96 \frac{\sigma}{\sqrt{n}},$
where 1.96 cuts probability 2.5% from the upper tail of the standard normal
distribution. (This confidence interval is based on the fact that
$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}}$ has a standard normal distribution.)
Notice that the theoretical standard error $\sigma/\sqrt{n}$ is used when $\sigma$ is known.
Transcribing from comments.
For final populations there exists a correction factor which will probably useful for this purpose.
$FPC=\sqrt{(N-n)/(N-1)}$,
and the standard error formula becomes
$SE=s/\sqrt{n} \cdot FPC$.
Best Answer
No, you will not have a "root-n" effect regardless of those things, since at least some standard errors do not scale with $\sqrt{n}$.
Many do -- quite possibly all the ones you will be likely to use -- but that's not all of them.
For things that do scale with $\sqrt{n}$ then you expect to halve the standard error by quadrupling sample size. So (at least if we're ignoring sampling variation in the estimate of $\sigma$), that's probably what you need.
One example of something that isn't proportional to $\frac{1}{\sqrt{n}}$ is the standard error of a kernel density estimate when the bandwidth is itself chosen as a function of $n$. [For some common choices of bandwidth formula the standard error goes down as $n^{-2/5}$ instead.]