Not a proof, but some thoughts on how to get empirical data for this problem and evidence for the optimal configuration:

Suppose we have a matrix $C$ whose rows are the unit vectors pointing in the directions of our cylinders. Then computing the Hausdorff distance amounts to finding, among unit vectors $v$ in space, the one which maximizes

$$\min_{c_i\in C}\frac{1}{\sqrt{1-||v\cdot c_i||^2}}$$

If we want to approximate this with a collection of random unit vectors given by rows in a matrix $V$, then this amounts to performing some simple element-wise arithmetic on the matrix $CV^T$, which means that it's quite fast to throw into any framework for doing matrix multiplications efficiently.

I initialized $1000$ random cylinder positions (forcing WLOG the first cylinder to point along $[1,0,0]$ and the second to be in the $xy$-plane), evaluating their score by taking the maximum radius along any of $100,000$ random vectors.

I then took the best of these initial candidates, made random small adjustments to some of the coordinates, and kept the change if it improved the score. (As the changes got smaller, I started re-scoring any candidates that looked like they were the best yet, since you're also putting a lot of selection on "drawing a lucky set of random vectors" and can get unrealistically good scores.)

The winner had cylinders pointed along the following vectors:

```
[ 1. 0. 0. ]
[ 0.49990595 -0.8660797 0. ]
[ 0.50013575 0.86594231 -0.00285279]
```

This seems like pretty strong evidence that the proposal in the original post is optimal.

With four cylinders, the winning candidate looked like:

```
[ 1. 0. 0. ]
[ 0.00114353 0.99999935 0. ]
[-0.7066738 0.70753928 -0.00054313]
[ 0.70754814 0.70600075 0.03063601]
```

This is very close to arranging the cylinders in a regular octagon in the $xy$-plane (and doesn't score better than doing so exactly).

With five cylinders:

```
[ 1. 0. 0. ]
[-0.80883904 -0.58803011 0. ]
[ 0.30972531 -0.9506074 -0.0203915 ]
[-0.80888608 0.58783778 -0.01224957]
[-0.30840091 -0.95123839 -0.00586572]
```

In general with $k$ cylinders, placing them at equally spaced intervals in a plane will acheive a Hausdorff distance of $\sec\left(\frac{\pi}{2k}\right)-1$, while placing $k-1$ of them equally spaced in the $xy$-plane and one of them vertically will acheive a distance of $\sqrt{1+\sin^2(\frac{\pi}{2(k-1)})}-1$, which is strictly worse for $k\ge 3$.