This reminds me of the notion of $(\epsilon, k)$-wise independence for random bit vectors. That is, given a set of $n$ random binary bits $X_i \sim \text{Bernoulli}\left(\frac{1}{2}\right)$, they are said to be $(\epsilon, k)$-wise independent if for any $S \subset [n], |S| = k,$ we have that $\left| \text{Pr}\left(\cap_{i \in S}\mathbb{1}\{X_i = 1\}\right) - 2^{-k} \right| \leq \epsilon$. This came up when I was crashing a course on probabilistic algorithms.

In your scenario, seeing as we are comparing the ratio between probabilities, we would achieve $k$-wise independence if the ratio $r := \frac{\text{Pr}\left(\cap_{i \in S}\mathbb{1}\{X_i = 1\}\right)}{2^{-k}} = 1$. Naturally the comparison to make then it would make sense to define this around the inequality $|r - 1| \leq \epsilon$, or for that matter try to bound the difference of log probabilities in an $\epsilon$-ball.

---EDIT---

I was thinking about this a little more, I think the independence relation we want to look at is indeed $r = 1 \iff \log\text{Pr}\left(\cap_{i \in S}\mathbb{1}\{X_i = 1\}\right) - \log2^{-k} = 0$.

In the same vein as almost $k$-wise independence, we would like to bound this difference in some $\epsilon$-ball, choosing $\epsilon < k\log\alpha$, since the inequality:

$$
\alpha^{-k} \leq \frac{\text{Pr}\left(\cap_{i \in S} X_i\right)}{\prod_{i \in S}\text{Pr}(X_i)} \leq \alpha^k
$$

implies the inequality in the case of random binary bits

$$
\left|\log\text{Pr}\left(\cap_{i \in S}\mathbb{1}\{X_i = 1\}\right) - \log2^{-k}\right| \leq k\log\alpha.
$$

Your particular presentation of probabilities generalizes this to finite collections of arbitrary random variables.

Some of the more popular resources on the topic:

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.437.3861&rep=rep1&type=pdf

http://people.csail.mit.edu/ningxie/papers/AAKMRX07.pdf