Ball-Box Theorem and Sequence of Distributions

Dealing with such a low regularity is a tricky business. However in dimension 3 you can get away with your set of assumptions.

First, if the distributions are uniformly Lipschitz and converge in $C^0$, then for every fixed $r$ the $r$-balls converge in the Hausdorff metric. This is easy to see if you choose coordinates $(x,y,z)$ in $\mathbb R^3$ so that the planes of the distributions are separated away from the $z$-axis. In this case every path $t\mapsto(x(t),y(t))$ in the $xy$-plane starting at $(0,0)$ uniquely determines a path $t\mapsto(x(t),y(t),z(t))$ tangent to the distribution and starting at $(0,0,0)$. The coordinate $z(t)$ obtained as the solution of an O.D.E., and these solutions converge since the coefficients are uniformly Lipschitz and converge in $C^0$.

However, the convergence of balls doest not mean the the full orbits converge. If the orthogonal component of the Lie bracket does not vanish, then the orbit is the entire space no matter what. Yet the limit distribution may be integrable and hence have two-dimensional orbits. There is nothing mysterious in that. The orbit is the set of points where the sub-Riemannian distance to the origin is finite. It can be finite for every $k$ but go to infinity as $k\to\infty$, so the limit orbit can be smaller. You can see a similar effect if you consider just the images of linear maps $\mathbb R\to\mathbb R$: for a map $x\mapsto \frac1k x$ the image is the entire line, but the limit map is zero.

As for the "thickness" of balls, they are bounded above in terms of the maximum of $([e,g],f)$ over a neighborhood. And from below in terms of the minimum of $([e,g],f)$. So if the bracket goes to 0 uniformly, you have an expected upper bound for the thickness. But if you want to control the thickness more tightly, you probably need to assume something to prevent fast oscillation of $([e,g],f)$ between zero and the maximum value (or between positive an negative).

You can prove the above mentioned bounds "by hand" if you choose a suitable coordinate system. For example, the one given by a local diffeomorphism $$ (x,y,z) \mapsto \exp(xe)\circ\exp(yg)\circ\exp(zf) . $$ Of course you may want to estimate the distortion of this coordinate system with respect to the original one. This is in some sense "inner working of the inverse function theorem", but you can do it directly by ODE analysis.

In these coordinates, one can move from $(x,y,z)$ to $(0,0,z)$ cheaply: just move time $-x$ along $e$, then time $-y$ along $g$. So you only need to control the changes of the $z$-coordinate as you move along the distribution. The vector field $e$ in these coordinates is the coordinate field $\partial/\partial x$, and $[e,g]=\partial g/\partial x$, so everything is indeed controlled by the 3rd coordinate og $[e,g]$.

I do not understand exactly what you mean by $\exp_x$, but I guess it is something like the diffeomorphism proposed by Sergei Ivanov. I want to stress out that the ball-box theorem holds only if the coordinates are privileged w.r.t. $(e_k,g_k)$, so you should check that the $exp_x$ you are using defines privileged coordinates for any $k$, in order to apply this theorem.

Regarding the question about $[e_k,g_k]\rightarrow [e,g]\in\text{span}(e,g)$, it is a quite delicate matter. Indeed, this is exactly what happens when we try to get uniform estimates for the sub-Riemannian balls on a sequence of points approaching a singular point (think for example of the Martinet distribution, which is of step 1 outside the plane $\{y=0\}$, where it is step 2, and take any sequence of points $(x_k,y_k,z_k)$ with $y_k>0$ but $y_k\rightarrow 0$). This problem has been treated in the paper

F. Jean, "Uniform Estimation of Sub-Riemannian Balls", Journal on Dynamical and Control Systems, vol. 7 (4), 2001 (http://www.ensta.fr/~fjean/jdcs03_07_01.pdf)

Also, in the notes on s-R geometry by Jean (http://arxiv.org/abs/1209.4387) you can find a somewhat more detailed discussion of privileged coordinates and their subtleties than in the Agrachev, Barilari, Boscain ones.