With a Bayesian method we could also consider $\bar{X} = \frac{1}{n} \sum_{k=1}^n X_k$ as the observed statistic and it has approximately a normal distribution if we assume that the values have finite variance and converges quickly.

So we could use the likelihood function $\mathcal{L}(\mu \vert \bar{X}) \approx \frac{1}{\sqrt{2 \pi \sigma^2/n}} \exp \left(\frac{(\bar X - \mu)^2}{2 \sigma^2/n} \right)$

Then we still need priors for $\sigma$ and $\mu$ but that is like any other Bayesian problem. The issue with the likelihood has been solved by assuming a normal distribution just like with the frequentist method.

With a Bayesian method we could also consider $\bar{X} = \frac{1}{n} \sum_{k=1}^n X_k$ as the observed statistic and it has approximately a normal distribution if we assume that the values have finite variance and converges quickly.

So we could use the likelihood function $\mathcal{L}(\mu \vert \bar{X}) \approx \frac{1}{\sqrt{2 \pi \sigma^2/n}} \exp \left(\frac{(\bar X - \mu)^2}{2 \sigma^2/n} \right)$

Then we still need priors for $\sigma$ and $\mu$ but that is like any other Bayesian problem. The issue with the likelihood has been solved by assuming a normal distribution just like with the frequentist method.

Related question: Would you say this is a trade off between frequentist and Bayesian stats?