Somewhere along the way I make an error and I can't seem to find where.
I mostly use central moments in the derivation so I will state them here for the sake of completeness: for $X \sim \mathcal{N}(\mu, \sigma^2)$, $$ \mathbb{E}(X - \mu)^2 = \sigma^2, \quad \mathbb{E}(X - \mu)^4 = 3\sigma^4, \quad \mathbb{V}(X - \mu)^2 = \mathbb{E}(X - \mu)^4 - (\mathbb{E}(X - \mu)^2)^2 = 3\sigma^4 - \sigma^4 = 2\sigma^4. $$
Firstly, it's easy to show that the estimator $\hat{\sigma}^2_0(\mathbf{X})$ is unbiased. Therefore, the information inequality (or equality for uniform minimum variance unbiasedness) takes the form $$ \mathbb{V}(\hat{\sigma}^2_0(\mathbf{X})) \geq \frac{1}{n I_1(\theta)}. $$ where $I_1(\theta)$ is the Fisher information number. Lets derive $I_1(\theta)$ first \begin{align} I_1(\theta) = \mathbb{E}(\frac{\partial}{\partial \theta} \log p(x, \theta))^2 & = \mathbb{E}(\frac{\partial}{\partial \theta}(-\frac{1}{2\sigma^2}(x - \mu)^2 - \log(\sqrt{2\pi}\sigma ))^2, \\ & = \mathbb{E}(\frac{1}{\sigma^3}(x-\mu)^2 - \frac{1}{\sigma})^2, \\ & = \mathbb{E}(\frac{(x-\mu)^2 - \sigma^2}{\sigma^3})^2, \\ & = \frac{\mathbb{E}(x-\mu)^4 - 2 \sigma^2 \mathbb{E}(x-\mu)^2 + \sigma^4}{\sigma^6}, \\ & = \frac{3\sigma^4 - 2\sigma^4 + \sigma^4}{\sigma^6}, \\ & = \frac{2}{\sigma^2}. \end{align} Now for the variance of the estimator, \begin{align} \mathbb{V}(\hat{\sigma}^2_0(\mathbf{X})) & = \mathbb{V}(n^{-1} \sum_{i=1}^n (X_i - \mu_0)^2) \\ & = n^{-2} \sum_{i=1}^n \mathbb{V}(X_i - \mu_0)^2 \\ & = n^{-2} n 2 \sigma^4 \\ & = n^{-1} 2\sigma^4. \end{align} Now the inequality takes the form $$ \frac{2\sigma^4}{n} \geq \frac{\sigma^2}{2n} $$ So what is missing is a square value of $\sigma$ and a factor of 2.
