As Valery Alexeez wrote
From Weibel, "An introduction to homological algebra", 2.3.13 on p.43 follow that $\mathcal{A}^I$ has enought projectives. But isnt clear if for a projective $P\in \mathcal{A}^I$ each $P(i)\in \mathcal{A}$ is projective as we need further.

(this is true if for any $i\in I$ the right Kan extention of the $i$-valutation $v(i): \mathcal{A}^I\to \mathcal{A}$ is exact, I dont know if this follow from the "$I$ is filtrant" hypothesis).

Alternatively:

From the book "Theory of Categories (BArry Mitchell) cor.7.6 p. 138, let $T_i: \mathcal{A}^I\to \mathcal{A}$ ($i\in I$) the $i$-valuation, and $S_i$ its left adjuction (the left Kan extention), now for a projective $P\in \mathcal{A}$ the object $S_i(P)(j),\ j\in I$ is a sum of copies of $P$ (see how the left KAn extention is maked, for example in the Weibel reference above) then is projective. THe above corollary asserts that projectives of $\mathcal{A}^I$ are objects of the form $\bigoplus_{i\in I}S_i(P_i)$ (where $P_i\in \mathcal{A}$ is a prjective) and all its retracts.

This ensure the existence a projective resolution of a diagram $(X_i)_i\in \mathcal{A}^I$ with projective arguments.