Why do better π-acceptor ligands cause smaller Δ(T) d-orbital splitting?
Since not many people know the ligand formazanate (I hadn’t heard of it before this question), let me quickly supply an image of what it looks like:


Scheme 1: General formazane structure.

Coodination happens from the lower two nitrogens, one of which is deprotonated (hence -ate) and both of which chelate the metal with $\mathrm{sp^2}$-type orbitals. The five formazane atoms form a &#x3c0; system, which can be extended by appropriate $\ce{R}$-groups. And one of these $\ce{R}$-groups is now modified to include an electron-withdrawing group.

I already previously gave a qualitative orbital scheme for a tetrahedral complex including &#x3c3; and &#x3c0; interactions as given below.


Figure 1: Qualitative MO-scheme of a tetrahedral $\mathrm{d^0}$ complex with &#x3c3; and &#x3c0; interactions. Originally created for this answer.

The &#x3c3; donor orbitals transform according to $\mathrm{a_1 + t_2}$, leaving $\mathrm{e + t_1 + t_2}$ for the &#x3c0; ligands. Unfortunately, the picture is a tad clouded, since both &#x3c3; and &#x3c0; interactions have a $\mathrm{t_2}$ contribution.

Adding an electron-withdrawing group to your ligand will leave the &#x3c3; interaction unchanged in first approximation, since mainly the ligand’s &#x3c0; system would be affected. The orbital energies of the &#x3c0; system would be reduced. Since we are talking about a metal with a low oxidation state, its orbitals are higher than drawn in the qualitative scheme (which was designed for permanganate). And thus also, the most important &#x3c0; interaction of the ligand will be, as you stated, accepting electrons into its &#x3c0; system.

The exact nature of any effect observed will depend on the original orbital energies. However, the $\mathrm{e}$ orbitals are only affected by the &#x3c0; interaction while the $\mathrm{t_2}$ orbitals are also affected by the &#x3c3; interaction. The magnitude of &#x3c0; interaction is expected to be the same for both groups of orbitals, since the start from the same value on each side. Thus, the difference between $\mathrm{e}$ and $\mathrm{t_2}$ stems mainly from &#x3c3; interactions which only affect $\mathrm{t_2}$ due to symmetry.

Because the &#x3c3; interaction starts with the ligand orbitals being much lower in energy than the metal orbitals, reducing their energy even more will result in a weaker interaction, therefore a lowering of the $\mathrm{t_2}$ level and thus a smaller $\Delta_\mathrm{T}$ or $10~\mathrm{Dq}$ value.