Note that the electric force between two charges is $$F_{\rm elec}=\frac{q_1q_2}{4\pi\epsilon_0 r^2},$$ the gravitational force is $$F_{\rm grav}=G\frac{M_1M_2}{r^2},$$ and finally the (maximal) force for two dipoles is $$F_{\rm mag}= \frac{3\mu_0 |\mathbf m_1|| \mathbf m_2|}{2\pi r^4},$$ where $\mathbf m_1,\mathbf m_2$ are the magnetic dipole moments.

On the same grounds as the standard electric and gravitational parameters, can just forget about the distance dependence and you just you define a standard magnetic parameter by multiplying the magnetic field constant by the dipole moment $\mathbf m$: $$\mu_{\rm mag}=\frac{\mu_0}{4\pi}|\mathbf m|\;\text{or}\;\frac{3\mu_0}{2\pi}|\mathbf m|$$.

Note that the electric force between two charges is $$F_{\rm elec}=\frac{q_1q_2}{4\pi\epsilon_0 r^2},$$ the gravitational force is $$F_{\rm grav}=G\frac{M_1M_2}{r^2},$$ and finally the (maximal) force for two dipoles is $$F_{\rm mag}= \frac{3\mu_0 |\mathbf m_1|| \mathbf m_2|}{2\pi r^4},$$ where $\mathbf m_1,\mathbf m_2$ are the magnetic dipole moments.

On the same grounds as the standard electric and gravitational parameters, can just forget about the distance dependence and you just you define a standard magnetic parameter by multiplying the magnetic field constant by the dipole moment $\mathbf m$: $$\mu_{\rm mag}=\frac{\mu_0}{4\pi}|\mathbf m|\;\text{or}\;\frac{3\mu_0}{2\pi}|\mathbf m|$$.

The constant does not matter much, just that it is proportional to its magnetic dipole moment. Some values: Mars has almost none magnetic moment, while Earth's is $8\times10^{22}$ A m${}^2$ while Jupiter has $1.5\times10^{27}$ A m$^{2}$. Source.