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.
Because $\sum_{i=1}^n a_i = 0$, $\sum_{i=1}^n a_i x_i =\sum_{i=1}^n a_i x_i -\bar x \sum_{i=1}^n a_i = \sum_{i=1}^n a_i (x_i-\bar x)$. So (4) can be written as
$$V(\tilde{\beta}) = \frac{\sigma^2 \sum_{i=1}^n a_i^2}{\left[\sum_{i=1}^n a_i x_i\right]^2}=\frac{\sigma^2 \sum_{i=1}^n a_i^2}{\left[\sum_{i=1}^n a_i (x_i - \bar x)\right]^2}$$
According to Cauchy-Schwarz Inequality, we have
$$\left[\sum_{i=1}^n a_i (x_i - \bar x)\right]^2 \le \sum_{i=1}^n (x_i-\bar x)^2\sum_{i=1}^n a_i^2$$
Divided by $\left[\sum_{i=1}^n a_i (x_i - \bar x)\right]^2 \sum_{i=1}^n (x_i-\bar x)^2$ on both sides, we get:
$$\frac{\sum_{i=1}^n a_i^2}{\left[\sum_{i=1}^n a_i (x_i-\bar x)\right]^2} \ge \frac 1 {\sum_{i=1}^n (x_i-\bar x)^2} = \frac {\sum_{i=1}^n (x_i-\bar x)^2}{\left(\sum_{i=1}^n (x_i-\bar x)^2\right)^2}$$
So $V(\hat \beta) \le V(\tilde \beta)$
Best Answer
This answer extends the sample mean case studied in Asymptotic distribution of $\sqrt{n}(\hat{\sigma}_{1}^{2}-\sigma^2)$ to the linear regression setup.
Let us assume an iid sample from a linear model $y_i=x_i'\beta+\epsilon_i$ with first four conditional error moments $\{E(\epsilon_i^p|x_i)\}_{p=1}^4=(0,\sigma^2,\mu_3,\mu_4)'$. Note $$\varsigma^2:=Var(\epsilon_i^2|x_i) = E(\epsilon_i^4|x_i)-[E(\epsilon_i^2|x_i)]^2=\mu_4-\sigma^4$$ and assume $0<\varsigma^2 <\infty$. That is, we require higher-moment assumptions not required for, e.g., "mere" consistency of the OLS estimator. In addition to assuming homoskedasticity, we also assume what one might call "homokurtosis", i.e., $Var(\epsilon_i^2|x_i)$ does not depend on $x_i$.
We may then add and subtract to get \begin{align*} s^2 & = \frac{1}{n}\sum_{i=1}^n (y_i-x_i'\hat\beta)^2\\ & = \frac{1}{n}\sum_{i=1}^n ((y_i- x_i'\beta) + (x_i'\hat\beta-x_i'\beta))^2\\ & = \frac{1}{n}\sum_{i=1}^n (\epsilon_i + (\hat\beta-\beta)'x_i))^2\\ & = \frac{1}{n}\sum_{i=1}^n \epsilon_i^2 + 2 (\hat\beta-\beta)'\frac{1}{n} \sum_{i=1}^nx_i\epsilon_i + \frac{1}{n}\sum_{i=1}^n[(\hat\beta-\beta)'x_i]^2\\ & = \underbrace{\frac{1}{n}\sum_{i=1}^n \epsilon_i^2}_{A} + \underbrace{2 (\hat\beta-\beta)'\frac{1}{n} \sum_{i=1}^nx_i\epsilon_i}_B+ \underbrace{(\hat\beta-\beta)'\frac{1}{n}\sum_{i=1}^nx_ix_i'(\hat\beta-\beta)}_C \end{align*}
$B$ and $C$ are asymptotically negligible under suitable assumptions: The OLS estimator converges at rate $\sqrt{n}$, i.e., $\hat\beta-\beta=O_p(n^{-1/2})$. Under predeterminedness (implied by the exogeneity condition you state), a WLLN gives $$ \frac{1}{n} \sum_{i=1}^nx_i\epsilon_i\to_pE(x_i\epsilon_i)=0, $$ i.e., $$\frac{1}{n} \sum_{i=1}^nx_i\epsilon_i=o_p(1)$$ Thus, \begin{align*} B & = O_p(n^{-1/2})o_p(1)=o_p(n^{-1/2}) \end{align*} Similarly, with existing second moments of the regressors we have $$ \frac{1}{n}\sum_{i=1}^nx_ix_i'\to_pE(x_ix_i')<\infty, $$ so that $$ \frac{1}{n}\sum_{i=1}^nx_ix_i'=O_p(1), $$ Then, $$ C=O_p(n^{-1/2})O_p(1)O_p(n^{-1/2})=O_p(n^{-1})=o_p(n^{-1/2}) $$
Hence, we may focus on the key term $A$ now. By assumption we have $0<\varsigma^2 <\infty$ and hence, by an application of the central limit theorem with $E(\epsilon_i^2)=\sigma^2$ we obtain that $$\sqrt{n}(s^2-\sigma^2)=\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n \epsilon_i^2+o_p(n^{-1/2})-\sigma^2\right) \stackrel{d}{\to}\mathcal{N}(0,\varsigma^2).$$
Writing $$ \varsigma^2=\mu_4-\sigma^4, $$ highlights the dependence of the asymptotic distribution on higher moments. E.g., for normally distributed errors, $\mu_4=3\sigma^4$, so that $\varsigma^2=2\sigma^4$.
Two illustrations based on normal and Laplace errors, depending on what you comment in:
The picture is for the Laplace case: