I did not see this term before, but my guess is that it is used in opposition to the term "infinitely exchangeable". The latest is necessary for the deFinetti representation theorem. As you have seen this term in the context of permutation tests, thst seems reasonable, since permutation tests do not need the stronger infinite exchangeability, as it only uses probability calculations for the fixed sample size $n$, no asymptotic calculations which might need the stronger concept. See Can someone explain the concept of 'exchangeability'? for more discussion on exchangeability.
As for the question on permutation tests for regression parameters, have a look at the book A Chronicle of
Permutation Statistical
Methods
1920–2000, and Beyond by Kenneth J. Berry, Janis E. Johnston, and
Paul W. Mielke Jr. Its section 5.32 Kennedy–Cade and Multiple Regression gives some permutation tests for multiple regression, see this paper.
More formally, we say that the random vector $X=(X_1,X_2, \dotsc, X_n)$ is exchangeable (or weakly exchangeable, if my guess is correct) if all permutations of the components of $X$ have the same distribution, that is,
$$
(X_{\pi 1},X_{\pi 2}, \dotsc, X_{\pi n})
$$
have the same distribution as $X$ for all permutations $\pi$ of $\{1,2,\dotsc,n\}$. This condition is not enough for the validity of deFinetti's representation theorem. For that we need infinite exchangeability, meaning that for all $m>n$ there exist a random vector $Y$, say, of dimension $m-n$, such that the distribution of $(X,Y)$ is exchangeable in the above sense.
An example where this extension is not possible is an equicorrelated multinormal vector with, say, expectation vector 0 and covariance matrix with all diagonal elements 1 and all off-diagonal elements the value $\rho <0$. If this multinormal vector has dimension $n$, there is an inequality
$$
\rho \ge -\frac1{n-1}
$$
Say, $n=3$ and $\rho=-0.1$. The above inequality gives $\rho \ge -0.5$, so is fulfilled. But then we try to extend with the new vector $Y$, we have the inequality
$$
-0.1 \ge -\frac1{m-1}
$$
which solved give $m \le 11$. So we can extend $X$ with new components until length 11, but then it stops, so $X$ is not infinitely exchangeable. But if my guess is correct, we can say that $X$ is weakly exchangeable. And, in the context of permutation tests, that looks reasonable. (In the multaivariate normal equicorrelated case, the condition necessary for infinite exchangeability is that $\rho \ge 0$).
Another example of an weakly exchangeable sequence which is not infinitely exchangeable is a multinomial random vector $X_1, X_2, \dotsc, X_k$ with probability vector $(1/k, 1/k, \dotsc, 1/k)$, since the covariances are negative.
