Measuring the magnetic field strength on the surface of a neodymium magnet (without a gaussmeter)? As the title states, I need to measure the magnetic field strength on the surface of a neodymium magnet but I do not have access to a gaussmeter.
Initially, I tried placing the magnet next to a compass so that the magnet's magnetic field is acting on the compass needle in a perpendicular direction to Earth's magnetic field. Then, using the angle the needle is displaced from pointing north (and earth's known magnetic field), the magnet's magnetic field can be worked out. Of course, however, the magnet's magnetic field is much stronger than that of the earth, so, the compass needle pointed directly at the magnet.
To solve this, I decided to back the magnet up far enough that the compass needle is displaced by a lesser amount. As I was not sure exactly how the distance between the magnet and the compass needle changes the magnetic field strength (perhaps someone can clear this up?), I took a number of measurements of the angle displacement at different distances.
Making a plot revealed that the magnet's magnetic field strength is inversely proportional to the cube of the distance from the compass needle. This concerned me because at distance 0 (the surface of the magnet), the magnetic field strength is "infinite." Of course, this is incorrect but I am not sure where I have gone wrong. Could someone point out the flaw in my understanding and suggest how the experiment could be modified to find the magnetic field strength on the surface of the magnet (or an alternate better experiment)?
 A: I'm not sure I can provide a full answer, but I figure I'll get the conversation started. 
Your magnet is essentially equivalent to a loop of wire with a given current (correct me if wrong). As such it would be expected that the Lorentz force will drop off $\propto \frac{1}{r}$. This is still likely imperfect, and I'm entirely unsure as to whether or not a compass could provide data without a laughably bad error.
My go-to for this would be to use induction. I would suspect (again correct me in the comments), that this could be derived from dropping the magnet through a solenoid and using an ammeter. If the coil was around the same size as the magnet, then you could probably calculate $\frac{\partial \vec{B}}{\partial t}$ and work from there.
A: You found correct correlations between deflection angle of compass and distance to your magnet. No error in your measurements. Monopole fields (such as gravity, electric charges, etc) drops in intensity with relation to distance as $\propto 1/r^2$. And dipole fields,- like electric charge $+-$ dipole or in this case - magnetic field (magnet with N-S poles),- drops in intensity by distance to field source like $\propto 1/r^3$. This is due to the fact that at the bigger distances- one pole cancels the effect of the other, so the net effect is that dipole field weakens faster due to superposition of these opposite poles.
If there would be some "magic" monopole magnetic source (with only one pole),- then you would get $\propto r^{-2}$. But as far as I know monopole magnets are just theoretical concept. And this can be perfectly understood why. In principle magnetic field is outcome of same orientation of spin between electrons. I.E. in magnetic material all electrons have spin oriented the same way. So that you can say that electron itself is like a "micro-magnet" with opposite poles. When bunch of these micro-magnets aligns the same - you get macro-magnet - material which magnetic field can be measured.
A: 
I do not have access to a Gaussmeter

It is highly likely that you have, as most mobile phones currently have magnetometers, and there are free applications to measure magnetic fields.
EDIT (July 27, 2022): However, do not place a strong magnet too close to your mobile phone.
A: Yes, on axis, the magnetic field of a neodymium magnet will approximate to to that of a dipole and vary as $r^{-3}$, where $r$ is the distance to the centre of the magnet.
This approximation is unlikely to work when $r$ is only a little bigger than the physical magnet - you should find the field is weaker than the $r^{-3}$ law.
