Consider $N$ i.i.d. samples $x_1, \dots, x_N$ from an unknown discrete distribution on $\{0,\dots, n\}$. We know that
$$\frac{1}{N-1} \sum_{i = 1}^N (x_i - m)^2$$
where $m$ is the sample mean, is an unbiased estimator for the variance.
But what confidence intervals can we give for this estimate?
On
You may use the asymptotic approximation. Specifically, there is enough structure to invoke the CLT, i.e., letting $S_n^2$ denote the sample variance ($n$ here denotes the sample size), \begin{align} \sqrt{n}(S_n^2-\sigma^2)&=\frac{1}{\sqrt{n}}\sum_{i=1}^n\left[(X_i-\mathsf{E}X_i)^2-\sigma^2\right]+o_p(1)\\ &\xrightarrow{d}N(0,\sigma^2(\kappa-1)), \end{align} where $\kappa:=\mu_4/\sigma^4$, $\mu_k$ is the $k$-th central moment of $X_1$, and $\sigma^2=\operatorname{Var}(X_1)$. Therefore, the asymptotic confidence interval for $\sigma^2$ (at nominal level $\alpha$) is of the form $$ S_n^2\pm z_{1-\frac{\alpha}{2}}S_n^2\sqrt{(K_n-1)/n}, $$ where $K_n$ is the sample kurtosis.
If the $X_i$ are i.i.d. normally distributed $N(\mu,\sigma^2)$ with $s^2=\frac{1}{N-1} \sum\limits_{i = 1}^N (x_i - \bar{x})^2$ then $$\frac{(N-1)s^2}{\sigma^2} \sim \chi^2_{N-1}$$
so, for example, if you want a $95\%$ confidence interval for the variance then you could use something like $$\left[ \frac{(N-1)s^2}{\chi^2_{0.025,N-1}}, \frac{(N-1)s^2}{\chi^2_{0.975,N-1}} \right]$$
You may need other approaches if the $X_i$ are not normally distributed; if your underlying distribution was Binomial with known $n$ and unknown $p$, the normal approximation may be good enough with large $N$, or you could take a more specific approach