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).
A lens does not have have one specific magnification, it depends on the positioning of the lens. When neglecting aberrations, the workings of a lens can be simplified with the following equation,
$$
\frac{1}{f} = \frac{1}{v} + \frac{1}{b},
$$
where $f$ is the focal length of the lens, $v$ is the distance from the object to the lens and $b$ the distance from the lens to the image of the object. This is demonstrated in the image below, including three principal rays (these only apply for thin lenses).
Here $\text{F}_1$ and $\text{F}_2$ are the two focal points of the lens, with $f_1$ and $f_2$ as their respective focal lengths (these are often equal to each other, which is also assumed in the first equation).
The resulting magnification, $M$, will be equal to the ratio between $h_1$ and $h_2$, which when expressed in terms of $v$ and $b$ looks as follows,
$$
M = \frac{v}{b}.
$$
When you have a lens with a given focal length then you have two equations with three unknown. So, when you want to calculate the magnification you would not have a unique solution. However manufactures probably want to add a label to their lenses which a layman can understand. For this they probably will use eyepiece magnification,
$$
M_e = \frac{250\ mm}{f}
$$
where the numerator is equal to the least distance of distinct vision, which is roughly 250 mm for a human with normal vision.
However if you do not know the focal length you can use the following equation,
$$
\frac{1}{f} = (n - 1) \left[\frac{1}{R_1} - \frac{1}{R_2} + \frac{(n - 1) d}{n R_1 R_2}\right],
$$
where $n$ is the refractive index of the lens material, $d$ the thickness of the lens, $R_1$ and $R_2$ the radius of curvature of the two sides of the lens.
Best Answer
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.