Learning a variance is hard.
It takes a (perhaps surprisingly) large number of samples to estimate a variance well in many cases. Below, I'll show the development for the "canonical" case of an i.i.d. normal sample.
Suppose $Y_i$, $i=1,\ldots,n$ are independent $\mathcal{N}(\mu, \sigma^2)$ random variables. We seek a $100(1-\alpha)\%$ confidence interval for the variance such that the width of the interval is $\rho s^2$, i.e., the width is $100\rho \%$ of the point estimate. For example, if $\rho = 1/2$, then the width of the CI is half the value of the point estimate, e.g., if $s^2 = 10$, then the CI would be something like $(8,\,13)$, having a width of 5. Note the asymmetry around the point estimate, as well. ($s^2$ is the unbiased estimator for the variance.)
"The" (rather, "a") confidence interval for $s^2$ is
$$
\frac{(n-1) s^2}{\chi_{(n-1)}^{2\;(1-\alpha/2)}} \leq \sigma^2 \leq \frac{(n-1) s^2}{\chi_{(n-1)}^{2\;(\alpha/2)}} \>,
$$
where $\chi_{(n-1)}^{2\;\beta}$ is the $\beta$ quantile of the chi-squared distribution with $n-1$ degrees of freedom. (This arises from the fact that $(n-1)s^2/\sigma^2$ is a pivotal quantity in a Gaussian setting.)
We want to minimize the width so that
$$
L(n) = \frac{(n-1) s^2}{\chi_{(n-1)}^{2\;(\alpha/2)}} - \frac{(n-1) s^2}{\chi_{(n-1)}^{2\;(1-\alpha/2)}} < \rho s^2 \>,
$$
so we are left to solve for $n$ such that
$$
(n-1) \left(\frac{1}{\chi_{(n-1)}^{2\;(\alpha/2)}} - \frac{1}{\chi_{(n-1)}^{2\;(1-\alpha/2)}} \right) < \rho .
$$
For the case of a 99% confidence interval, we get $n = 65$ for $\rho = 1$ and $n = 5321$ for $\rho = 0.1$. This last case yields an interval that is (still!) 10% as large as the point estimate of the variance.
If your chosen confidence level is less than 99%, then the same width interval will be obtained for a lower value of $n$. But, $n$ may still may be larger than you would have guessed.
A plot of the sample size $n$ versus the proportional width $\rho$ shows something that looks asymptotically linear on a log-log scale; in other words, a power-law--like relationship. We can estimate the power of this power-law relationship (crudely) as
$$
\hat{\alpha} \approx \frac{\log 0.1 - \log 1}{\log 5321 - \log 65} = \frac{-\log 10}{\log \frac{5231}{65}} \approx -0.525 ,
$$
which is, unfortunately, decidedly slow!
This is sort of the "canonical" case to give you a feel for how to go about the calculation. Based on your plots, your data don't look particularly normal; in particular, there is what appears to be noticeable skewness.
But, this should give you a ballpark idea of what to expect. Note that to answer your second question above, it is necessary to fix some confidence level first, which I've set to 99% in the development above for demonstration purposes.