John, your method for determining approximate lens focal lengths is a valid one, although since your eye will automatically accommodate to some extent, accuracy will be limited.
The apparent discrepancies follow from the rather obscure convention of labeling lens "Power" in units of Diopters, which are inverse meters.
The power P of an optic is defined as the inverse of the focal length f in meters:
$$ P = \frac{1}{f} $$
Thus a lens of power $-3$ has a focal length of $-1/3$ meter or $-33$ $cm$, not far from the $f= -30$ $cm$ you estimated. Similarly, a power of $-0.25$ Diopters corresponds to $-4$ $m$ or $-400$ $cm$, a very weak negative lens which would scarcely be noticeable when combined with a fairly high powered positive lens (magnifier).
Larger focal length lenses do have less optical power, as you have stated.
A single lens, however, does not form a telescope. You need two lenses, e.g. an objective lens and an eye lens with respective focal lengths $f_\text{eye}$ and $f_\text{obj}$. The magnification of a telescope constructed with these two lenses will have a magnification
$$ M = \left|\frac{f_\text{obj}}{f_\text{eye}}\right|$$
A larger objective focal length does result in larger magnification -- however, the individually higher optical power element in this telescope is actually the eye lens.
To achieve large magnification, you need either a really short focal length eye lens or a really long focal length objective. It is much more difficult to manufacture high quality short focal length lenses than it is to manufacture high quality long focal length lenses. This is because short focal lenses are more curved. In the thin lens approximation, the focal length of a lens of index $n$ in air with radii of curvature $R_1,R_2$ is given by:
$$ \frac{1}{f} = (n-1)\left( \frac{1}{R_1} - \frac{1}{R_2}\right)$$
As a lens focal length gets longer and longer, the lens becomes less and less curved. An infinite focal length lens would be a pane of glass, or a window.
Can you use a "window" as your objective lens to achieve infinite magnification? In principle, yes, however, this is not a practical solution primarily because the length of a two-lens telescope is $L = f_\text{obj} + f_\text{eye}$. In other words, such a telescope would need to be infinite in length.
Best Answer
You are correct that for a single lens the working distance would be the focal length. For compound lenses, like microscope objectives, you have to look at the entire optical system to figure out the working distance. The short answer to your question is that the focal length and working distance are not what you expect because the objective is a compound lens. The magnification for microscope objectives is confusing because to calculate the magnification you must know something about the tube lens that is used along with the objective.
I'm going to go through an example, for quantities I don't define explicitly here, please see Figures 9 and 12 in refrence 1.
Suppose you have two focusing lenses (f1, and f2) separated by a distance d were the first lens is a distance W from the object you are trying to image. Microscope objectives have to fit in a specific space so they have a required parfocal distance PD = d + W (I am assuming thin lenses).
The ray tracing optics for this lens system would be: \begin{equation} M= \left( \begin{array}{cc} M_{11} & M_{12} \\ M_{21} & M_{22} \\ \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ -\frac{1}{\text{f2}} & 1 \\ \end{array} \right).\left( \begin{array}{cc} 1 & PD-W \\ 0 & 1 \\ \end{array} \right).\left( \begin{array}{cc} 1 & 0 \\ -\frac{1}{\text{f1}} & 1 \\ \end{array} \right).\left( \begin{array}{cc} 1 & W \\ 0 & 1 \\ \end{array} \right) \end{equation}
Actual performing the math we get: \begin{equation} M= \left( \begin{array}{cc} 1-\frac{\text{PD}-W}{\text{f1}} & \text{PD}+\left(1-\frac{\text{PD}-W}{\text{f1}}\right) W-W \\ -\frac{1-\frac{\text{PD}-W}{\text{f2}}}{\text{f1}}-\frac{1}{\text{f2}} & -\frac{\text{PD}-W}{\text{f2}}+\left(-\frac{1-\frac{\text{PD}-W}{\text{f2}}}{\text{f1}}-\frac{1}{\text{f2}}\right) W+1 \\ \end{array} \right) \end{equation}
In order for this objective to image properly we set the requirement that M12 = 0. When we are happy that our objective correctly images we can figure out the effective focal length by calculating the M21 element.
Let's work an example. Suppose that PD = 45 mm, f1 = 11 mm, f2 = 16.5 mm. I mostly made these numbers up, but PD for microscope objectives is 45 - 60 mm. This is on purpose to make objectives interchangeable.
From the requirement that M12 = 0 we can calculate W: \begin{equation} M_{12} = W \left(1-\frac{\text{PD}-W}{\text{f1}}\right)+\text{PD}-W = 0 \\ \implies W = 19.15 mm, \end{equation} so the working distance is 19.15 mm.
Next we want to calculate the focal distance of this lens, which we can get from the M21 element: \begin{equation} M_{21} = -(1/f2) - (1 - (PD - W)/f2)/f1 = -0.0091 mm^{-1}, \end{equation} but focal lengths are usually written like $M_{21} = -1/f_{o}$ so the focal length of our compound lens system is $f_{o}$ = 110 mm. So very clearly, compound lenses do not behave like single lenses.
What about the magnification? To answer that question we have to look at the M matrix with all the numbers plugged in: \begin{equation} M= \left( \begin{array}{cc} -1.35037 & 0. \\ -0.00906831 & -0.740536 \\ \end{array} \right). \end{equation}
From inspection you'd expect the magnification of this lens to be 1.35 (the minus sign just means the image is upside down). This is where your confusion comes in: magnification for microscope objectives require knowledge of the tube lens. From this example of someone trying to sell you a microscope objective you can see that it clearly states that:
So for our example above using a 200 mm tube lens the magnification would be $M=f_{tube}/f_{objective} = 200 / 110 = 1.82$. Scrolling down the page of reference 2 you can see that the FL=100 mm lens has a magnification of 2.