How would a bayesian estimate a mean from a large sample? What would a bayesian do if she wanted to do inference for the mean with a large sample but has no idea of the underlying distributions?
A frequentist statitician would use the sample mean as a point estimate and CLT for the distribution of the estimator. All she has to assume is finite variance.
I have found Bernstein Von-Mises Theorem which states that given a random sample $X_1, \ldots, X_n$ and

*

*a prior $p(\mu)$ for the mean

*a distribution for the sample given the parameter $f_{x | \mu}$
Then if the sample size is sufficiently large the posterior distribution of the mean given the sample is approximately normal, i.e
$$ \mu_{|_{X_1, \ldots X_n}} \approx \mathcal{N}\left( \hat{\mu}_{ml}\, ; \, \frac{1}{I(\mu_0) n}\right)$$
Where the mean is the maximum likelihood estimator and $I(\mu_0)$ is Fisher information number for the true mean.
In the case where we have no knowledge of $f_{x | \mu}$ there is the problem that  maximum likelihood nor fisher information number can be calculated.  But what can be done? Becuase even if we do not know it the distribution exists and it will be approximately normal.

Intentions:
Some buisness woman has to take an action based on the true sign of the mean. A decision has to be made so she will use the sign of the sample mean but she needs some measure of the potential cost that arise in case of making the wrong decistion.
So, if the sample mean is positive things like
$$ P\left( \mu < 0 | X_1 \ldots X_n \right) = \int_{-\infty}^0 f_{\mu |x_1 \ldots x_n}( t) \, dt.$$  or
$$ E \left( \mu I_{(-\infty, 0)}| X_1 \ldots X_n \right)  = \int_{-\infty}^0 t f_{\mu |x_1 \ldots x_n}( t) \, dt.$$
would make a lot of (buisness) sense for her.
In the frequentist context she is stuck with affirmations of the kind:
With $95\%$ confidence the mean is greater than a certain value.

You seem to be asking about a nonparametric estimator for the mean.
First, let's make it clear: for Bayesian statistics, you always need to make distributional assumptions. You can proceed as suggested by Sextus Empiricus (+1), but this does assume Gaussian distribution. If you really didn't want to make any assumptions, in practice, you would probably just estimate the arithmetic mean.
But let's try coming up with a nonparametric solution. One thing that comes to my mind is Bayesian bootstrap also described by Rasmus Bååth who provided code example. With Bayesian bootstrap, you would assume the Dirichlet-uniform distribution for the probabilities, resample with replacement the datapoints and evaluate the statistic, so arithmetic mean on those samples. But in such a case, to approximate the frequentist estimator, you would use the same estimator on the samples to find their distribution. Not very helpful, isn't it?
Let's start again. The definition of expected value is
$$
E[X] = \int x \, f(x)\, dx
$$
The problem is that you don't know the probability densities $f(x)$. With the parametric model, you would solve it by finding a parametric distribution for $f$. In the frequentist setting, we would estimate the expected value using the arithmetic average with $\phi_i$ weights equal to the empirical probabilities
$$
\widehat{E[X]} = \sum_{i=1}^N x_i \, \phi_i 
$$
Same as with Bayesian bootstrap, you could assume a prior Dirichlet-uniform prior for the $\phi_i$ weights, sampling the weights, calculating the weighted average, and repeating this many times to find the distribution of the estimates. If you think about it, It's the most trivial case of the Bayesian bootstrap for a weighted statistic. Yes, this makes a number of unreasonable assumptions like approximating continuous distribution with a discrete one. It's also not necessarily very useful, but if you insist on a Bayesian nonparametric estimator, that's a possibility. As you can see from the example below, it gives the same results as the frequentist estimator and standard bootstrap.
set.seed(42)

N <- 500
X <- rnorm(N, 53, 37)
mean(x)
## [1] 51.88829

R <- 5000
mean.boot <- replicate(R, mean(sample(X, replace=TRUE)))
summary(mean.boot)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   46.41   50.81   51.90   51.91   53.03   57.88 

# weighted.mean itself normalizes the weights
wmean.boot <- replicate(R, weighted.mean(X, w=rexp(N, 1)))
summary(wmean.boot)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   47.09   50.82   51.93   51.91   52.99   57.65 

par(mfrow=c(1, 2))

hist(mean.boot, 100, freq=FALSE, main="Standard bootstrap")
curve(dnorm(x, mean(X), sd(X)/sqrt(N)), 45, 60, add=T, col="red", lw=2)

hist(wmean.boot, 100, freq=FALSE, main="Sampled weights")
curve(dnorm(x, mean(X), sd(X)/sqrt(N)), 45, 60, add=T, col="red", lw=2)


In the end, it's the same as the frequentist estimator, so you could just use the frequentist estimator.
