When we calculate Fisher's LSD test, why do we use Mean Square of Error, $\sigma_e$, which is the variance of all groups (pooled variance), as a variance of each individual group mean? In the denominator it's:
$$\sqrt{{{\sigma_e} \over {n_1}}+ {{\sigma_e} \over {n_2}}}$$
Why don't we use group variance instead of MSE? Why don't we use a t-test just like the procedure of testing differences betweeen two sample means for independent samples?
If you are assuming (as usual in a one-way ANOVA) that all $g$ treatment groups have the same variance, then Fisher's LSD has greater power if you use data from all $g$ groups to get the standard error for the difference.
Recall that Fisher's LSD is to be used only if the F-test in the ANOVA has found that there are significant differences among the group means.
This method of finding the standard error is what distinguishes Fisher's LSD from doing ${g \choose 2}$ tw0-sample tests to try to find the pattern of differences among the group means.
Note: Of course, if group means truly have quite different population means, the F-test in the ANOVA does not have the anticipated distribution, and Fisher's LSD does not work properly either. In that case, there is a separate-variances Welch-style ANOVA procedure that is appropriate, and LSD would not be used as the method of multiple comparisons among group means. Bonferroni seems to be the multiple comparison method of choice. Some software implements the separate-variances ANOVA (for example, R) and other software does not (for example, Minitab).