I don't know if you can quantify the confidence level of the z-values, but you can certainly give a smaller confidence value for your estimates of the mean.
Read the wiki page on standard error. Standard error differs from standard deviation in that standard error is the amount of variation in the parameter estimates. In this case, the standard error should be calculated as
$\hat \sigma_\mu = \frac{\hat \sigma_x}{\sqrt{N}}$
where $N$ is the sample size. As such, your estimates for the mean will have much tighter bounds when you calculate a confidence interval (e.g., $95\% CI = \hat{\mu} \pm 1.96\hat{\sigma}_\mu$). As $N$ increases, the bounds will get tighter.
Now, if you're dead set on finding an estimate for the z-score, we can find error bounds for it. Assuming you know the population mean $\mu$ and variance $\sigma$ of the null distribution, we can define a new random variable, $Z$:
$Z = \frac{X- \mu}{\sigma}$
We can find the standard error or error depending on which you'd prefer. It sounds like you're more interested in finding the standard error, so let's look at
$ \hat{\mu}_Z = \frac{ \hat{\mu}_X- \mu}{\sigma} $
$Var(\hat{\mu}_Z)
= Var(\frac{\hat{\mu}_X- \mu}{\sigma})
= \frac{Var(\hat{\mu}_X) - Var(\mu)}{\sigma^2}
= \frac{Var(\hat{\mu}_X)}{\sigma^2}$
Since $Var(\hat{\mu}_X) = \frac{\hat{\sigma}_X}{\sqrt{N}}$, the standard error for a mean z-score is
$Var(\hat{\mu}_Z) = \frac{\hat{\sigma}_X}{\sqrt{N}\sigma}$
You can generalize to two samples as well; it all comes down to basic variance calculations. Bear in mind this is from calculating a z-score from a sample mean, you can also derive things from working with Z. Bottom line is divide by sigma for most of these z normalizations.
