Call the schemes to be intersected $(X_i)$, where $X_i$ has pure codimension $r_i$ in ${\mathbb P}^n$. Let $R = \sum r_i$. (Edited so as not to restrict to $R = n$ unnecessarily.)
Definitely, every component of the intersection has codimension at most $n$$R$. If the codimensions are all exactly $n$$R$, and the schemes being intersected are Cohen-Macaulay, then the product of the degrees = the degree of the intersection ( = the sum of the degrees of its primary components).
Non-example: let $X$ be the projective completion of a random plane through the origin in $A^4$, and $Y$ the projective completion of the union of two other random planes through the origin (so, not Cohen-Macaulay). Then $X \cap Y$ is a triple point, not a double point as one might hope $(deg\ X = 1,deg\ Y = 2)$. The basic issue is that if we think about intersecting $Y$ first with a $3$-plane $X' \supset X$, we get a union of two lines plus an embedded point we should throw away before we go all the way down to $X$. Then the intersection of $X$ picks up a point for each line in $Y \cap X'$, which is good, but also the embedded point, which is a failure of codimensions adding up.
In this non-example $r_1 = r_2 = 2$, $n = R = 4$.