Sure, you can generalize the mean free path to a different number of dimensions. But first, let's understand the derivation in 3D.
A particle will collide with any other particle that it comes within a distance $d$ of. So if it moves a length $\ell$, it will collide if there is another particle in a cylindrical volume $\pi d^2 \ell$. Call this the volume swept out by the particle.
Imagine a region of volume $V$ filled with gas, and let $\bar{v}$ be the mean speed of the particles of that gas in the rest frame of the entire volume. Then, in the reference frame of one of the gas particles, the other particles have an average speed of $\sqrt{2}\bar{v}$. (See this derivation of the factor of $\sqrt{2}$, and note the presence of $\bar{v}$ instead of $v_\text{rms}$ - see this question for details on that)
In the reference frame of that one particle, the total volume swept out by all the other particles in a time $\Delta t$ is $N\pi d^2 \sqrt{2}\bar{v}\Delta t$, where $N$ is the number of particles.
The probability that the chosen at-rest particle experiences an interaction during $\Delta t$ is equal to the fraction of the total volume ($V$) swept out, namely
$$P(\text{int.}) = \frac{N\pi d^2 \sqrt{2}\bar{v}\Delta t}{V}$$
To calculate the mean free time $\tau$, I technically should find the probability distribution of interaction times and compute its mean. But to make the calculation simple, I'll take advantage of a handy coincidence (which I am not justifying here): $\tau$ happens to be equal to the time after which this probability would reach 1 if it increased at a fixed rate over time. So I can replace $\Delta t$ with $\tau$ and $P(\text{int.})$ with $1$, and I get
$$\tau = \frac{V}{N\pi d^2 \sqrt{2}\bar{v}}$$
The mean free path is then given by
$$\lambda = \bar{v}\tau = \frac{V}{\sqrt{2}N\pi d^2}$$
If you assume the particles follow the ideal gas law, you can replace $\frac{V}{N} = \frac{k_B T}{p}$ and recover the formula from Wikipedia.
To modify this to 2D, we just need to change step 1 and follow the argument from there.
Instead of a particle sweeping out a volume $\pi d^2\ell$, it sweeps out an area $2d\ell$.
No change
The total area swept out by all $N$ particles is then $Nd\sqrt{8}\bar{v}\Delta t$
The probability is equal to the fraction of the total area,
$$P(\text{int.}) = \frac{Nd \sqrt{8}\bar{v}\Delta t}{A}$$
The mean free time is again the time after which this probability would reach 1,
$$\tau = \frac{A}{Nd\sqrt{8}\bar{v}}$$
and the mean free path is
$$\lambda = \bar{v}\tau = \frac{A}{\sqrt{8}Nd}$$
Bear in mind that in 2D, the ideal gas law would be modified; it would have area instead of volume, and you would have to use a 2D version of pressure, which would be force per unit length, not per unit area. It may be easier to just work with $\lambda = \frac{A}{\sqrt{8}Nd}$ directly, since you can just set $A = L^2$ if you know the box side length.
Gravity makes molecules gradually accelerate downwards. Neglecting collisions, the molecules closer to the earth would thus be (on average) moving faster.
You cannot neglect collisions, at least not in the part of the atmosphere where the atmosphere acts like a gas. Collisions remain important until you get to the exobase. Above the exobase, the atmosphere is so rare that collisions can be ignored. Modeling the exosphere is messy because now you have to worry about solar flares, the Earth's magnetic field, and a varying distribution of components (e.g., the exosphere is dominated by hydrogen).
Collisions are extremely important in the thick part of the atmosphere (up to 120 km or so). At sea level, the mean free path (average distance between collisions) in the atmosphere is less than 1/10 of a micrometer. At 20 kilometers altitude, the mean free path is about one micrometer. By the time you get to 120 kilometers, the mean free path grows to about a meter. Collisions remain important until you get to the exobase.
Temperature in the atmosphere follows a complex profile. The highest temperatures are in the two highest layers of the atmosphere, the thermosphere and exosphere.
Your second argument suffers many of the same problems as your first. You cannot ignore collisions. Collisions are an essential part of what make a gas a gas. Things are a bit murky in the one part of the atmosphere, the exosphere, where collisions can be ignored. There is one easy way to model the exosphere: It's essentially a vacuum. Getting past that easy model is non-trivial. Even the thermosphere (the next layer down, where the space station orbits) is problematic to model. The atmosphere is still thin enough in the thermosphere that it doesn't act quite like a gas.
It's the mesosphere on down where the atmosphere acts like a gas. That the components are constant colliding with one another is what makes pressure in the lower atmosphere equal to the weight of all the stuff above. This isn't necessarily true in the upper atmosphere.
Best Answer
the collision sets the time scale over which the particle can travel freely on average. Based on this picture, semiclassical, you can use the relation you quoted to calculate the Diffusion constant.