Uncountable Mittag-Leffler condition?

Question

Let $(X_\alpha)_{\alpha <\kappa}$ be an inverse system of abelian groups.

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If $\kappa = \omega$ (or by extension if $\kappa$ is of countable cofinality), then the Mittag-Leffler condition is a criterion guaranteeing that $\varprojlim^1_{\alpha <\kappa} X_\alpha = 0$. Namely, one says that $(X_n)_{n < \omega}$ is Mittag-Leffler if for every $n < \omega$, there exists $n' \geq n$ such that for every $n'' \geq n'$, one has $\operatorname{Im}(X_{n''} \to X_n) = \operatorname{Im}(X_{n'} \to X_n)$. In other words, $(X_n)_{n<\omega}$ is Mittag-Leffler if "$\operatorname{Im}(X_m \to X_n)$ stabilizes for $m \geq n$ sufficiently large". Two special cases are worth mentioning:

1. If every map $X_m \to X_n$ is an epimorphism, then $(X_n)_{n<\omega}$ is Mittag-Leffler.

2. If every map $X_m \to X_n$ with $m > n$ is zero, then $(X_n)_{n<\omega}$ is Mittag-Leffler.

And to reiterate: the point of the Mittag-Leffler condition is the following theorem of Grothendieck:

Theorem: If $(X_n)_{n<\omega}$ is Mittag-Leffler, then $\varprojlim^1_{n<\omega} X_n = 0$.

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Now consider the case where $\kappa$ is an uncountable regular cardinal. In this case, the Mittag-Leffler condition as formulated above is not sufficient to guarantee that $\varprojlim^1_{\alpha<\kappa} X_\alpha = 0$. Moreover, in this case it's possible to have $\varprojlim^n_{\alpha<\kappa} (X_\alpha) \neq 0$ for $n \geq 2$. Nevertheless, an analog of (1) above is available. That is Neeman shows (Triangulated Categories, Appendix A) the following:

1. (bis) If $X_\alpha \to \varprojlim_{\beta<\alpha} X_\beta$ is an epimorphism for each $\alpha <\kappa$, then $\varprojlim^n_{\alpha<\kappa} = 0$ for all $n \geq 1$.

(When $\alpha = \beta+1$ is a successor, this reduces to $X_{\beta+1} \to X_\beta$ being an epimorphism as in (1); when $\alpha$ is a limit, this is an additional condition.) In fact, Neeman simply calls the condition of (1 bis) "Mittag-Leffler". I'd resist this choice of terminology, because unlike in the countable case, (1 bis) does not contain as a special case the following analog of (2):

2. (bis) if $X_\alpha \to X_\beta$ is zero for all $\kappa > \alpha > \beta$, then $\varprojlim^1_{\alpha <\kappa} X_\alpha = 0$.

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Question: Let $\kappa$ be an uncountable regular cardinal, and let $(X_\alpha)_{\alpha < \kappa}$ be an inverse system of abelian groups (or even vector spaces). Is there a natural condition, analogous to the Mittag-Leffler condition when $\kappa = \omega$, which one can put on $(X_\alpha)_{\alpha<\kappa}$ ensuring that $\varprojlim^1_{\alpha < \kappa} X_\alpha = 0$?

In particular, is there such a condition which contains both (1 bis) and (2 bis) as special cases?