David,
At first sight I think you might be right about this. Personally, I try to avoid using Weyl's inequality in this form, but rather some statement of the following form: If $|S_N| \geq \epsilon N$, and if $\epsilon > N^{-c}$, then there is some $q \leq \epsilon^{-C}$ such that the fractional part of $q\theta$ is at most $\epsilon^{-C}/N^d$.
In other words: if the exponential sum is large then $\theta$ is very close to a rational with denominator $q$.
You can prove this variant in much the same way, that is to say by repeated squaring. I don't think the inequality is typically stated in this way because the form you stated leads to better bounds in Waring's problem, where the factor of $(\log N)^C$ isn't very important. However, as you point out, it does seem to be important when talking about the equidistribution result.
Let me try to be a little more specific about how to prove this ``log free'' variant of Weyl's inequality that I've mentioned. Presumably there is a reference in the literature. However you can start from the presentation that I give on pages 59-60 of these notes
http://www.dpmms.cam.ac.uk/~bjg23/AddNumTheory/chap3.ps
At some point one obtains many $h_1, h_2, \dots, h_d$ for which $\Vert \theta h_1,\dots, h_d \Vert$ is small. At this point it is standard to invoke the divisor function estimate to show that there are in fact many $n$ for which $\Vert \theta n \Vert$ is small. However in doing this one loses an $N^{\epsilon}$ (it's worse than $\log^C N$ - are you sure you've quoted Gowers accurately?). To avoid losing it, let $S$ be the set of all $h_1\dots h_d$ mentioned above. Then $\Vert \theta (s_1 + s_2 + \dots + s_m) \Vert$ is small for all choices of $s_1,\dots, s_m \in S$, and one can argue* that for big enough $m$ this set of sums of $S$ is really big (i.e. there is no loss of $N^{\epsilon}$.)
*The key point is that for large enough $m$, the number of representations of any $n \in [X^d, 2X^d]$ as a sum of $m$ things of the form $h_1 \dots h_d$, $h_i \sim X$ is bounded by $C X^{d(m-1)}$. The usual proof would have $m = 1$, where this statement is actually false. The problem is, I think I'd need to use the Hardy-Littlewood method (which uses Weyl's inequality, but only the weaker form with the $N^{\epsilon}$) to prove this statement! Little surprise that you don't find this argument in textbooks then.
Actually, I'd be very interested to see a decent reference for all this.