To the extent you can assume your process is a white stationary ergodic process, the variance of the mean is the variance of the process divided by N where N is the number of samples.
Assuming that you can estimate the variance from your sample using
$$\sigma_x = \frac{1}{N-1}\sum_{i=1}^N(x_i-\mu_x)^2$$
NOTE: This is the unbiased estimator since the variance is being estimated from a sample of N elements using a mean that is estimated from the sample itself. (see Zwillinger, D. (Ed.). CRC Standard Mathematical Tables and Formulae. Boca Raton, FL: CRC Press, 1995.). Otherwise if the mean was known (which it isn't) there would be a $\frac{1}{N}$ in front of the summation above.
You should then be able to derive confidence intervals on the mean itself with the variance of the mean as $\frac{\sigma_x}{N}$:.
This is seen in the expression for the mean and variance of $Y$ where $Y$ is the average of $X$ over $N$ samples, as follows where $\mu_x$ is the mean of x and $\sigma_x$ is the variance of x:
\begin{align}
E(Y) &= E\left[\frac{1}{N}\left(X_1+X_2+ \ldots +X_N\right)\right] = \frac{1}{N}E\left[(X_1+X_2+ \ldots +X_N)\right]\\
\textrm{Var}(Y) &= \textrm{Var}\left[\frac{1}{N}\left(X_1+X_2+ \ldots +X_N\right)\right] = \frac{1}{N^2}\textrm{Var}\left(X_1+X_2+ \ldots+X_N\right)
\end{align}
\begin{align}
E\left[(X_1+X_2+ \ldots +X_N)\right] &= E[X_1]+E[X_2]+ \ldots E[X_N] = N \mu_x\\
\textrm{Var}(X_1+X_2+ \ldots +X_N) &= \textrm{Var}(X_1)+\textrm{Var}(X_2) + \ldots \textrm{Var}(X_N) = N\sigma_x^2
\end{align}
Therefore
\begin{align}
E(Y) &= \frac{1}{N}E[(X_1+X_2+ \ldots +X_N)] =\frac{1}{N}N\mu_x = \mu_x\quad\text{and}\\
\textrm{Var}(Y) &= \frac{1}{N^2}\textrm{Var}(X_1+X_2+ \ldots +X_N) = \frac{1}{N^2}N\sigma^2 = \frac{\sigma^2}{N}
\end{align}
How to make use of the Autocorrelation Function
Note that this reduction by N is valid as long as the sequence is white. Once samples are correlated, there will be no further reduction in the variance of the estimate. You can therefore make use of your autocorrelation to determine the number of samples that will reduce the variance. I do not have an exact calculation for this, but to provide a rough order magnitude I would use the number of samples that results in the normalized autocorrelation dropping to below 0.5. For example, in your data the total number of samples is 30,000. If the autocorrelion immediately dropped below 0.5 after just one sample, then all samples are independent and your variance estimate would be the $\sigma_x/30,000$. However if the autocorrelation does not drop below 0.5 until 100 samples, then the variance estimate would be $\sigma_x/300$.
