I found some lecture notes from the course I asked about. I will relate the core idea here, as best I can.
First, I am going to assume any reader is familiar with the standard error of the sample mean: $$se(\mu) =\frac{\sigma}{\sqrt n}$$
Second, from here we get that the standard error of the sample variance for any distribution is: $$se(\sigma^2)=\sqrt{\frac 1 n (\mu_4-\frac{n-3}{n-1}\sigma^4)}$$
where $\mu_4$ is the fourth central moment: $E(X-\mu)^4$.
For a normal distribution $\mu_4 = 3\sigma^4$ so this simplifies to
$$se(\sigma^2)=\frac{\sqrt 2 \sigma^2}{\sqrt{n-1}}$$
Lastly, use the delta method to derive an approximation of the standard error of a transformed parameter with a known standard error. Here the parameter with the known standard error is variance, and the transformation is $g(\theta)=\sqrt \theta $. The link in the second point uses the delta method for just this purpose, and arrive at a result of
$$se(\sigma)\approx\frac 1 {2\sigma}se(\sigma^2)$$
Substitute in the value of $se(\sigma^2)$ and we get
$$se(\sigma)\approx\frac \sigma {\sqrt {2(n-1)}}$$
So the conclusion is that the standard error of the sample standard deviation is smaller than the standard error of the sample mean by a factor of $\frac 1 {\sqrt 2}$, for large n, for a normal distribution, and so is in this (very limited) sense, easier to estimate.
So the result isn't as general as what my memory suggested, but interesting nonetheless.
PS: the result holds for any distribution and sample size where kurtosis is small enough. Specifically when
$$\frac{\mu_4}{\sigma^4} < 4 + \frac{n-3}{n-1}$$