Tolerance and confidence in the minimum multi-variate effective sample size (minESS) The minimum multi-variate effective sample size (minESS) is defined in the R package mcmcse (where the function is implemented) as:

the minimum effective sample size required for a specified relative tolerance level

It depends on the number of free parameters being sampled by the MCMC algorithm, and on two other parameters: $\epsilon$, and $\alpha$, defined as:

  
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*$\epsilon$: tolerance level.
  
*$\alpha$: confidence level.
  

In this answer the above parameters are described as:

  
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*$\epsilon$: the precision. $\epsilon$ is the fraction of error you want the Monte Carlo to be in comparison to the posterior error. This is similar to the margin of error idea when doing sample size calculations in the classical setting.
  
*$\alpha$: the level for constructing confidence intervals.
  

This answer also describes $\alpha$ as the "confidence in your estimate" and $\epsilon$ as the "Monte Carlo error".
The meaning of $(\alpha\,, \epsilon)$ is still not entirely clear to me. Could you describe these parameters as basically as possible (ie: aimed at non-statisticians)?
 A: The minESS function is trying to calculate the minimum effective samples needed for a certain estimation problem. To recognize the interpretation of $\epsilon$ and $\alpha$, let me first go over the sample size calculations for a one sample mean.
Suppose $X_1, X_2, \dots, X_n$ is a sample from a distribution with mean $\mu$. The estimator for $\mu$ is $\bar{X}_n$, the average which is approximately normally distributed with variance $\sigma^2/n$ for large $n$. Now suppose in order to be happy with your estimate of $\mu$, you know that it must be within $1$ unit of $\mu$. Of course, due to randomness you can't guarantee $\bar{X}_n$ is within 1 units of $\mu$ all of the time, so instead you are happy to settle for $\bar{X}_n$ within 1 unit of $\mu$ 95% of the time on average. That is, if you were to make a confidence interval for $\mu$, you would want the 95% confidence interval to be of width $E = 2$ units (1 on each side).
So you need $n$ so that the variance $\sigma^2/n$ is small enough to allow that halfwidth. Since the halfwidth for a $z$-test is $z_{1-\alpha/2} \sigma/\sqrt{n}$, we want
$$\dfrac{z_{1-\alpha/2} \sigma}{\sqrt{n} } \leq E\,$$
 we get the require sample size is
$$n > \left(\dfrac{z_{1-\alpha/2}\, \sigma}{ E} \right)^2\,. $$
Now let's extend it to multiple dimensions. Suppose each $X_1, X_2, \dots X_n$ is a vector in $p$-dimensions and in order to capture cross-correlations we are creating multivariate confidence regions instead of individual confidence intervals for each component. This becomes more difficult because if $p$ is large then we may need to set a "half-width" for each component which will be exhaustive. In addition, all the components of $X_1$ may not be in the same units, so the half-width will really lose any interpretation.
Instead, an idea similar to "half-width" is used where we want an estimate $\bar{X}_n$ (which is a vector) so that the variability in the estimator is an $\epsilon$th fraction compared to the variability in the distribution from which it is sampled. To give some intuition on this, below is a plot from this thesis (Page 10)

On the left you have a target distribution whose standard deviation $SD = 5$ indicated by the grey area. The black area is the $\epsilon * SD$ where $\epsilon = .2$. So for the left plot we want the "half-width" of the confidence region to be smaller than the black box. However, on the right plot, you have a distribution with a smaller variance, $SD = 2$. A black box exactly the same width of the left plot would be much too wide, but  a relative idea, $\epsilon *SD$ would scale the black box appropriately down to a good size relative to that of $SD$.
In this way $\epsilon$ restricts the size of the confidence region relative to the inherent variability of the underlying distribution. $\alpha$ serves the same purpose as before since nothing is guaranteed due to randomness, a confidence level is needed. Now the way this is carried out (following notation in the link you provided), we get the required sample size when
$$\text{Volume of Confidence region of $\bar{X}_n$}^{1/p} \leq \epsilon |\Lambda|^{1/2p}\,,$$
notice that this similar to the univariate case where the volume of a confidence region has to bounded by some quantity. Now, the volume of a confidence region is $ n^{-1/2} c_{\alpha,p}  |\Sigma|^{1/2p}$, where $c_{\alpha,p}$ is a constant that depends on $\alpha$ and $p$. Moving things around,
$$mESS \geq c_{\alpha,p}\,, $$
with $c_{\alpha, p}$ being provided here.
