The general asymptotic result for the asymptotic distribution of the sample variance is (see this post)

$$\sqrt n(\hat v - v) \xrightarrow{d} N\left(0,\mu_4 - v^2\right)$$

where here, I have used the notation $v\equiv \sigma^2$ to avoid later confusion with squares, and where $\mu_4 = \mathrm{E}\left((X_i -\mu)^4\right)$. Therefore by the continuous mapping theorem

$$\frac {n(\hat v - v)^2}{\mu_4 - v^2} \xrightarrow{d} \chi^2_1 $$

Then, accepting the approximation,

$$P\left(\frac {n(\hat v - v)^2}{\mu_4 - v^2}\leq \chi^2_{1,1-a}\right)=1-a$$

The term in the parenthesis will give us a quadratic equation in $v$ that will include the unknown term $\mu_4$. Accepting a further approximation, we can estimate this from the sample. Then we will obtain

$$P\left(Av^2 + Bv +\Gamma\leq 0 \right)=1-a$$

The roots of the polynomial are

$$v^*_{1,2}= \frac {-B \pm \sqrt {B^2 -4A\Gamma}}{2A}$$

and our $1-a$ confidence interval for the population variance will be

$$\max\Big\{0,\min\{v^*_{1,2}\}\Big\}\leq \sigma^2 \leq \max\{v^*_{1,2}\}$$

since the probability that the quadratic polynomial is smaller than zero, equals (in our case, where $A>0$) the probability that the population variance lies in between the roots of the polynomial.

## Monte Carlo Study

For clarity, denote $\chi^2_{1,1-a}\equiv z$.

A little algebra gives us that

$$A = n+z, \;\;\ B = -2n\hat v,\;\; \Gamma = n\hat v^2 -z \hat \mu_4$$

which leads to

$$v^*_{1,2}= \frac {n\hat v \pm \sqrt {nz(\hat \mu_4-\hat v^2)+z^2\hat \mu_4}}{n+z}$$

For $a=0.05$ we have $\chi^2_{1,1-a}\equiv z = 3.84$

I generated $10,000$ samples each of size $n=100$ from a Gamma distribution with shape parameter $k=3$ and scale parameter $\theta = 2$. The true mean is $\mu = 6$, and the true variance is $v=\sigma^2 =12$.

**Results:**

The sample distribution of the sample variance had a long road ahead to become normal, but this is to be expected for the small sample size chosen. Its average value though was $11.88$, pretty close to the true value.

The estimation bound was smaller than the true variance, in $1,456$ samples, while the lower bound was greater than the true variance only $17$ times. So the true value was missed by the $CI$ in $14.73$% of the samples, mostly due to undershooting, giving a confidence level of $85$%, which is a $~10$ percentage points worsening from the nominal confidence level of $95$%.

On average the lower bound was $7.20$, while on average the upper bound was $15.68$.
The average length of the CI was $8.47$. Its minimum length was $2.56$ while its maximum length was $34.52$.

You are missing a lot of context here (presumably taken from linear regression) so I will focus solely on the asymptotic distribution:$^\dagger$

$$\sqrt N (\sigma^2 - \hat{\sigma}^2) \overset{\text{Dist}}{\rightarrow} \text{N}(0, 2 \sigma^4). \quad \quad$$

The left-hand-side here is a scaled version of the estimation error, which is affected by the size $N$. As $N \rightarrow \infty$ we obtain convergence in distribution to the distribution shown on the right-hand-side, which does not depend on $N$. This is sensible because although the distribution of the scaled estimation error should depend on $N$, its *limit* should not.

It is possible to re-frame the asymptotic result as an approximating distribution that becomes more and more accurate in the limit. (Indeed, this is the main value of an asymptotic distribution.) If $N$ is large we have:

$$\ \ \quad \sigma^2 - \hat{\sigma}^2 \overset{\text{Approx}}{\sim} \text{N} \bigg( 0, \frac{2 \sigma^4}{N} \bigg).$$

As you can see from this approximating distribution, the distribution of the estimation error is centred around zero (reflecting an unbiased estimator). As $N$ becomes larger, the (approximate) distribution of the estimation error has a lower variance, so it tends to get smaller. This gives us some useful consistency properties for the estimator, and it accords with our intuition that estimator becomes more accurate as we get more data.

$^\dagger$ Note that this notation is shorthand for the following formal mathematical statement:

$$\quad \quad \lim_{N \rightarrow \infty} \mathbb{P} \Big( \sqrt N (\sigma_p^2 - \hat \sigma^2) \leqslant \epsilon \Big) = \Phi \bigg( \frac{\epsilon}{\sqrt{2} \sigma^2} \bigg)
\quad \quad \quad \text{for all } \epsilon \in \mathbb{R}.$$

where the function $\Phi$ is the cumulative distribution function for the standard normal distribution.

## Best Answer

To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.

We also have $\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$. The first component converges in distribution to a Normal, the second convergres in probability to zero. Then by Slutsky's theorem the product converges in probability to zero,

$$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.