It could be the case that the macroscopic and the microscopic motions do "interact", but since the macroscopic wind is on such a large scale compared to the average kinetic energy of individual molecules, any system in which we use the average kinetic energy of the molecules is so small that the wind is effectively constant, and so it could be ignored.
Yes, this is -in a certain way- the right answer.
Temperature is proportional to the average kinetic energy, thus it is proportional to the average molecular speed squared. The exact formula for an ideal gas is
$$\langle v \rangle = \sqrt{\frac{2RT}{M}}$$
where $M$ is the molecular mass (the molecular mass of air is $\simeq 29$ g/mol).
If you have a volume of "still" air at temperature $T=25$°C ($298$ K), the molecules in such a volume will be moving randomly at an average speed of 467 m/s.
If you superimpose to such a motion a translation in some direction, the average speed registered by a stationary thermometer will be higher. But how fast can such a motion be?
The highest wind speed ever registered on Earth was registered on Barrow Island during the passage of the Cyclone Olivia (1996), and it was 113 m/s. Average wind speeds are far below this extreme values; you can see from this table that the order of magnitude is 10 MPH, i.e. 4.5 m/s.
So the presence of "ordinary" wind can increase the average molecular speed in one spatial direction of $\simeq 1\%$. I specified "in one spatial direction" because wind usually blows in a certain specific direction.
Let's take a reference frame such that the direction of the wind is the $x$ direction. If the other two components of the molecular speed remain unchanged and the speed in the $x$ direction increases of $1\%$, and if we assume that the three components where initially on average equal, we can roughly estimate the variation in $\langle v \rangle$ to be
$$
\frac{v'-v}{v} = \frac{\sqrt{v_x'^2+v_y^2+v_z^2}-\sqrt{v_x^2+v_y^2+v_z^2}}{\sqrt{v_x^2+v_y^2+v_z^2}} = \frac{\sqrt{(1.01 u)^2 + 2 u^2}-\sqrt{3u^2}}{\sqrt{3u^2}} \simeq 0.0033 = 0.33\%
$$
Since, from the previous equation,
$$T = \frac{M \langle v \rangle^2}{2R}$$
we see that the measured temperature will increase of $\simeq 0.67\%$ (i.e. $T'/T\simeq1.0067$), so if it was initially $25$°C $=298$K it will become $300$K = $27$°C.
So even in thi highly idealized situation the increase will not be very noticeable. In real life, we have at least two additional things to take into account:
Air is not an ideal gas. The most important factor is that air usually contains a non negligible amount of water vapor, the presence of which can greatly change its thermodynamic properties
In the atmosphere, wind is actually a large mass of air moving from regions of higher pressure to regions of lower pressure. Such a mass of air can be colder than the surrounding air, so the presence of wind can actually be linked to a decrease in temperature. The discussion becomes much more complicated in this cases, and a full understanding of it would require some knowledge of atmospheric science.