Years ago (2004) I was writing some notes for myself which included a section on some parts of Kelley's paper; looking at them now, I don't immediately see why things I omitted as "obvious" are obvious, but I reproduce what I wrote in case it is helpful. (I may try to come back and fill in more details later if required.)
The following definitions are made for subsets of $2^\Omega$, I trust that they would be equivalent to the definitions you gave above if one represents the given Boolean algebra $A$ as a Boolean sub-algebra of a suitable $2^\Omega$.
If $U_1,\dots, U_m$ are subsets of $\Omega$ (not necessarily distinct), and $\eta \in [0,1]$, we say the $m$-tuple ${\bf U}:=(U_1,\dots, U_m)$ is $\eta$-intersecting if
$$\sum_{i=1}^m 1_{U_i} \leq \eta\ m 1_\Omega $$
and denote the least such $\eta$ by $\iota({\bf U})$. Note that if $k\cdot {\bf U}$ denotes ${\bf U}$ repeated $k$ times then $\iota(k\cdot {\bf U})=\iota({\bf U})$.
For ${\mathcal C}\subseteq 2^{\Omega}$ we define $\iota({\mathcal C})$ to be $\inf_{\bf U} \iota({\bf U})$ where the infimum is over all finite tuples ${\bf U}=(U_1,\dots, U_m)$ where $U_i\in {\mathcal C}$ for all $i$.
The following is what I have written in my old notes, without further justification/calculation. It might be explained/motivated by things in Kelley's paper but I do not have a copy to hand right now.
Think of $2^{\Omega}$ as the Hamming cube sitting inside $\ell_\infty^{|\Omega|}$, and ${\mathcal C}$ as a set of vertices (=extreme points of $2^{\Omega}$). Then $\iota({\mathcal C})$ is the "inradius", with respect to the $\ell_\infty$-norm, of the convex hull of ${\mathcal C}\cup\{{\bf 0}\}$. More precisely, $\iota({\mathcal C}) = \inf\{ \Vert x\Vert_\infty \colon x\in {\rm conv}({\mathcal C})\}$.
Example.
Let $\Omega=\{1,2,3,4\}$ and let ${\mathcal C}$ be the following subset of $\ell_\infty^4$:
$$ {\mathcal C}=\{ (1,0,0,0), (0,1,1,0), (0,1,0,1), (0,0,1,1)\}. $$
An element of the convex hull is
$$ {\bf x} = (\lambda_1, \lambda_2+\lambda_3, \lambda_2+\lambda_4, \lambda_3+\lambda_4) $$
where $\lambda_i\in [0,1]$ for all $i$ and $\sum_i \lambda_i= 1$. Then $x_2+x_3+x_4=2(1-\lambda_1)$, so
$$ \Vert {\bf x}\Vert_\infty \geq \max(|x_1|, \tfrac{1}{3} |x_2+x_3+x_4| ) =\max\{ \lambda_1, \tfrac{2}{3}(1-\lambda_1)\ \colon 0\leq\lambda_1\leq 1\} \qquad(*)$$
By drawing the lines $y=t$, $y=\frac{2}{3}(1-t)$ one sees that the RHS of $(*)$ is always greater than the y-coordinate of the point where these two lines meet; this intersection point is $(2/5, 2/5)$. This argument shows that $\Vert{\bf x}\vert_\infty \geq 2/5$, and by taking $\lambda_1=2/5$ and $\lambda_2=\lambda_3=\lambda_4=1/5$ we see that this lower bound can be attained at some ${\bf x}\in {\rm conv}({\mathcal C})$.