Dear Wildfyr,
Don’t be frustrated because this seems a simple transformation but it is not simple at all.
1). “…The reaction is driven by the amine attacking the alkenyl chloride, then being deprotonated…”
Theoretically and if so, you don’t need an additional base at all; because the second molecule of diphenylamine will deprotonate the intermediate, diphenyliminium chloride. But this is not feasible because a borderline base (diphenylamine) cannot displace a hard base (Cl-).
2). By adding a base, you form the diphenyliminium anion, which is a strong and hard base that can diplace the chlorine atom by an ordinary SN2 substitution. But in that case, the second molecule of diphenylamine is not necessary.
3). However a hard base (K2CO3, MeO-, LiOH, etc.) will react the alkenyl chloride, in preference and not with the diphenylamine.
4). Except if using a tertiary amine as a base, which is a borderline base that cannot displace the chlorine atom but it can deprotonate diphenylamine, in equilibrium. But in this case, the second molecule of diphenylamine is necessary, in order to deprotonate the intermediately formed, diphenyliminium chloride.
Consequently, the desired product could be obtained:
1). Either by reacting 2 moles diphenylamine with 1 mole of alkenyl bromide or alkenyl iodide that are borderline and soft bases, respectively and in the absence of base.
2). Or by using a strong tertiary amine (e.g. DBU) in adequate amounts that are higher than catalytic ones, together with 2 moles diphenylamine and 1 mole of alkenyl chloride (or better, alkenyl bromide if possible).
Good luck!