Reed switch sensitivity/Strong magnets for 6 ft detection
Assuming you are considering the idea of a permanent magnet and not considering an electro-magnet, then the following reasoning can be applied:



Let's take a sensitive compass and allow it to align with the local magnetic field of Earth. In the US, this will vary a little. But \$B_\text{EARTH}\approx 20\:\mu\text{T}\$ would be a reasonable assumption in many places here. Let's assume that this is a variable and that you can find it in a book, somewhere, regarding your location on Earth. For now, just for conversational purposes, let's use the figure I just gave as usefully illustrative.

We can estimate the magnetic dipole moment of an iron-based ferromagnetic bar magnet, where it is made almost entirely of iron atoms. There are four theoretical ways (in a single atom, anyway) that are sources for the atomic "currents" that can generate magnetic fields:


Electrons "orbiting" the nucleus.
Electrons spinning on their own axes.
Proton charges "orbiting" inside the nucleus.
Proton charges spinning on their own axes.


The nuclear proton charges don't account for much, as it turns out. There are a variety of reasons, some depending on the fact that angular momentum is quantized and some other reasoning. But suffice it that it is the electron that dominates -- and it does so by a factor of almost 2000. So we need to focus on the first two, not the last two.

For an electron orbiting a nucleus, the angular momentum is quantized and must be an integer multiple of Planck's constant, \$h\$, divided by \$2\,\pi\$... aka \$\hbar\$. The electron spin is always a half-integer of \$\hbar\$. So assuming one quantum of angular momentum (N=1), we have \$L=\hbar\$ and therefore:

$$\mu=\frac12\frac{C_e}{m_e}\hbar=\frac12\cdot\frac{1.6\times10^{-19} \:\text{C}}{9\times 10^{-31} \:\text{kg}}\cdot 1.05\times 10^{-34}\:\text{J}\cdot\text{s}\approx 1\times10^{-23}\:\text{A}\cdot \text{m}^2$$

per charge in an atom.

If we assume that there is one charge per atom of iron in a bar magnet made of, say, \$100\:\text{g}\$ of iron, we would  find the following:

$$\begin{align*}
N&amp;=\frac{100\:\text{g}}{56\:\frac{\text{g}}{\text{mol}}}\cdot6.02\times 10^{23} \frac{\text{atoms}}{\text{mol}}\approx 1.075\times 10^{24} \:\text{atoms}\\\\\therefore\\\\
\mu&amp;=N\cdot \mu_\text{atom}=1.075\times 10^{24} \:\text{atoms}\:\cdot 1\times10^{-23}\:\frac{\text{A}\cdot \text{m}^2}{\text{atom}}\\\\&amp;=10.75\:\text{A}\cdot \text{m}^2
\end{align*}$$

In practice (and this is from experience with iron magnets) we'd probably actually find something closer to \$\mu=5\:\text{A}\cdot \text{m}^2\$. (This isn't bad. It's actually amazing that such a simple assumption can get so close to measured values.)

Now, let's take a quick look at neodymium magnets. Here, you find, "This magnetic energy value is about 18 times greater than "ordinary" ferrite magnets by volume and 12 times by mass." So for these magnets, particularly the more powerful ones, you may find the value to be about \$12\times\$ stronger. (I'm not sure, from experience, that this is accurate -- but let's take it as a given for the following analysis.) Therefore, a neodymium magnet weighing \$100\:\text{g}\$ may yield \$\mu=60\:\text{A}\cdot \text{m}^2\$.

From here, we can compute the deflection of the compass needle. For this given powerful magnet, at a distance of 6 feet, we can find the deflection as:

$$\theta=\tan^{-1}\left(\frac{\frac{2\,\mu_0\cdot \mu}{4\,\pi\,r^3}}{B_\text{EARTH}}\right)$$

If you plug in the values (such as \$\frac{\mu_0}{4\pi}=1\times 10^{-7}\:\frac{\text{T}\cdot\text{m}}{\text{A}}\$) and your distance as \$r\approx 1.83 \:\text{m}\$, then you will find the compass needle deflection to be about \$\frac12^\circ\$. This might be detectable, depending on how you manage your detection system for this small motion.

You could increase the mass of your neodymium magnet to improve this result, increasing the angle of deflection. So, if you could find a fairly massive neodymium magnet, you might find that you could detect it's presence more easily.

Note that here I've assumed some of the strongest permanent magnets readily available. So keep that in mind. But at least this provides a quantitative (and experience-based) result that answers your question. Your local magnetic field from the Earth matters, too, in this calculation. Keep that in mind, as well. But this, at least, demonstrates that it isn't beyond comprehension that you could detect the presence of a magnet at that distance. It's not easy, and there are some requirements about the magnet itself, but it's not impossible, either.

But at least, now, you have a way of quantitatively estimating the required mass of the magnet and its construction to yield a result you need.



P.S.
If you want to "go nuts," please look at this article. Here, you can find a new technique that would greatly increase the sensitivity of a compass needle. This could also greatly improve the sensitivity and substantially lower the mass of the permanent magnets you can detect at that distance.