The "mi" column is the chi-squared value, so you can find its probability using pchisq(). As Ed said, they are 1-df tests, so the critical value using alpha = 5% is 3.84 (without adjusting for multiple testing). You should control for the number of tests you are doing:
library(lavaan)
example(modindices)
MIs <- modindices(fit)
MIs$pvalue <- pchisq(MIs$mi, df = 1, lower.tail = FALSE) # naïve p value
## possibly save a subset of rows you are interested in, then...
MIs$bonf.p <- p.adjust(MIs$pvalue, method = "bonferroni") # keeps familywise Type I errors below alpha level
MIs$FDR.p <- p.adjust(MIs$pvalue, method = "fdr") # keeps false discovery rate below alpha level
MIs
A smarter, less overly conservative way to do so would be to consider ahead of time which modification indices you would be willing to free for theoretically justifiable reasons. Then you can save that subset of rows of the modindices() output, and the adjustments to p values will not be so conservative as to lose all power. Forethought about your tests is also a way to prevent inadvertently letting the data make decisions for you, which has been shown to consistently fail to lead researchers to a better-fitting model that also generalizes to new samples from the same population.
Terrence D. Jorgensen
Postdoctoral Researcher, Methods and Statistics
Research Institute for Child Development and Education, the University of Amsterdam