This is a good question, but at first blush it is hard to answer. This is because there is always an ambiguity in where you put the meaningful definitions - you can put it in the force constant, or you can put it in the units for the magnetic charges.
For magnetic monopoles, the first place one naturally turns to is to the magnetic Gauss law, which would be modified to the form $\nabla\cdot\mathbf B\propto \rho_\mathrm{m}$ ... except that you don't really have a way to fix that proportionality constant. By symmetry with respect to the electric Gauss law ($\nabla\cdot\mathbf E=\rho_\mathrm e/\varepsilon_0$) you'd hope that would be $\mu_0$, but that's some hardy guessing.
What you do want, very much, is for magnetic monopoles to have the same relationship to magnetic dipoles as electric monopoles have with electric dipoles - and magnetic dipoles we know how to handle.
If you take this as your basis, then, you can postulate* a magnetic field of the form
$$
\mathbf B=\frac{k_1q_\mathrm{m}}{r^2}\hat{\mathbf r}, \tag 1
$$
with some as-yet-to-be-determined constant $k_1$, for a magnetic monopole of magnetic charge $q_\mathrm{m}$. The key physical input is requiring a pair of opposite magnetic charges a distance $d$ apart to have a magnetic dipole moment of
$$
m=q_\mathrm{m}d,
$$
as in the dipole case, and this fixes the units of the magnetic charge: you know that this magnetic dipole (made from two opposite magnetic monopoles) must be exactly equivalent to a standard magnetic dipole: a current $I$ in a loop of area $A$, with magnetic dipole moment $m=IA$. This requires you to have $[q_\mathrm{m}]=\mathrm{A\,m}$.
From here you can get the dimensionality of the magnetic field constant:
$$
[k_1]
=[Br^2/q_\mathrm{m}]
=[BL/I]
=\left[\frac{\mu_0I}{L}\frac{L}{I}\right]
=[\mu_0]
=\mathrm{N/A^2}.
$$
However, it doesn't tell you the numeric value of this constant, for which you need to go back to your initial physical input - the equivalence of opposite pairs of magnetic monopoles with current loops. In particular, if each magnetic monopole has a magnetic field as in $(1)$, then it follows that a pair of them, with opposite magnetic charges and separated by a distance $d$ along an axis $\hat{\mathbf u}$, must have the magnetic field
$$
\mathbf B=\frac{k_1q_\mathrm{m}d}{r^3}\left(3(\hat{\mathbf u}\cdot\hat{\mathbf r})\hat{\mathbf r}-\hat{\mathbf u}\right).
$$
This contrasts with the magnetic field of a current-loop magnetic dipole with magnetic dipole moment $\mathbf m$
$$
\mathbf B=\frac{\mu_0/4\pi}{r^3}\left(3({\mathbf m}\cdot\hat{\mathbf r})\hat{\mathbf r}-{\mathbf m}\right),
$$
which needs to be equivalent under the identification $\mathbf m=q_\mathrm{m} d\:\hat{\mathbf u}$, and this completely fixes the magnetic-field constant at
$$
k_1=\frac{\mu_0}{4\pi},
$$
i.e.
$$
\mathbf B=\frac{\mu_0}{4\pi}\frac{q_\mathrm{m}}{r^2}\hat{\mathbf r}
$$
for a point magnetic monopole of magnetic charge $q_\mathrm{m}$. Similarly, this fixes the magnetic Gauss law to the naive
$$
\nabla\cdot\mathbf B=\mu_0\rho_\mathrm m
$$
(where $\rho_\mathrm m$ is of course the volumetric density of magnetic charge).
OK, so that's a lot of work - and we're nowhere near the force that you asked about. The reason for this is that we can only get out of the formalism what we put in, and we have only specified how magnetic monopoles should produce magnetic fields, but not how they should react to them. To get that from the formalism, we need to give it more information, and here again we are constrained in that a opposed-point-monopoles magnetic dipole needs to feel exactly the same force that a current-loop magnetic dipole of the same magnetic moment does.
Similarly to the above, we can postulate* that an external magnetic field $\mathbf B$ will produce a force
$$
\mathbf F=k_2q_\mathrm m\mathbf B
$$
on a point magnetic monopole of magnetic charge $q_\mathrm m$, and see what happens. Given this postulate, the same algebra that worked for electric dipoles implies that if you have two magnetic monopoles of opposite magnetic charges $q_\mathrm m$ a distance $d$ apart along the unit vector $\hat{\mathbf u}$, then the torque on them exerted by an external uniform magnetic field $\mathbf B$ will be
$$
\boldsymbol{\tau} = (k_2q_\mathrm md\:\hat{\mathbf u})\times\mathbf B,
$$
which contrasts with $\boldsymbol{\tau} = \mathbf m\times\mathbf B$ for a usual current-loop magnetic dipole of magnetic dipole moment $\mathbf m$. This equivalence then forces our second constant to be
$$
k_2=1.
$$
Similarly, you can check that this force will give identical expressions for the force on a magnetic dipole in a non-uniform external magnetic field $\mathbf B(\mathbf r)$, regardless of whether it is made up from opposing magnetic monopoles or from a small current loop.
With this, then, we can just connect the two main laws - how magnetic monopoles produce magnetic fields and how they react to them - to get the answer we're after. If you have two magnetic monopoles of magnetic charges $q_{\mathrm m,1}$ and $q_{\mathrm m,2}$, separated by a distance $r$ along the separation vector $\hat{\mathbf{r}}_{1\to2}$ pointing from $1$ to $2$, then the magnetic force exerted by magnetic monopole $1$ on magnetic monopole $2$ is
$$
\mathbf F=\frac{\mu_0}{4\pi} \frac{q_{\mathrm{m},1}q_{\mathrm{m},2}}{r^2}\hat{\mathbf{r}}_{1\to2}.
$$
As expected, like poles repel each other (e.g. two point north poles repel).
Here the magnetic charges are measured in ampere meters, and the magnetic charges can be calibrated by observing the interaction with a moving electric charge: if you have a point magnetic monopole of magnetic charge $q_\mathrm{m}$ and an electric charge $q_\mathrm{e}$ separated by a distance $r$ along the unit separation vector $\hat{\mathbf{r}}_\mathrm{m\to e}$, with the electric charge moving at velocity $\mathbf{v}_\mathrm{e}$, then the electric charge will be subject to a force
$$
\mathbf F=\frac{\mu_0}{4\pi} \frac{q_{\mathrm{m}}q_\mathrm{e}}{r^2} \mathbf{v}_\mathrm{e} \times \hat{\mathbf{r}}_\mathrm{m\to e}.
$$
This gives you an 'anchor' for the unit of magnetic charge - it's not free-floating, and it's completely tied to the unit of electric charge.
* Note that these are also additional impositions on the form of the produced magnetic field and the response to external fields. However, both postulates are reasonable things to assume: if the relations were substantially different, then we wouldn't want to call those objects magnetic monopoles. Moreover, it should be essentially possible to get to those forms using very little more than symmetry considerations.
In early times only static electricity and magnetism were known. No connection between these phenomena was suspected. Indeed, static electric and magnetic fields are unconnected. Then the attention shifted to varying electric and magnetic fields. Electricity and magnetism were found to be connected and Faraday's and Ampère's laws were formulated. So experimentally, the electromagnetic unification was almost complete. Then Maxwell added the displacement current and the theoretical unification was complete, enabling the prediction of EM waves and the identification of light as an EM wave.
Why then are we still using E and B as physical quantities? That is because the Lorentz force is written in terms of E and B and is gauge invariant. This led to the idea that all EM quantities should be gauge invariant and thus be expressible in E and B. L.V. Lorenz introduced what later became the four vector potential (V,$\mathbf A$). However at the time it was understood that all EM phenomena should involve just E and B and Maxwell formulated his theory in terms of E and B.
Are E and B then not unified? In my opinion the four potential unifies E and B. The potential is used in classical and quantum mechanics and you will have a hard time using E and B in its place. Also the Aharonov-Bohm effect cannot be explained by the B field. Nevertheless, due to the concept of gauge invariance it is believed that the potential is not physical. Therefor you could argue that the unification is incomplete. In my opinion the potential actually is entirely physical and unifies electromagnetism. I published a paper on this, which can be accessed on arxiv.org.
Best Answer
Both of these follow from desirable properties of this hypothetical magnetic charge, namely:
All three of these properties hold for electric charges. The last one may not be as familiar, but it basically works as follows: if we have a positive electric charge moving at constant velocity, it generates a magnetic field in addition to its electric field. A second positive electric charge moving parallel to the first one will therefore experience a magnetic force, and if you work out the directions, this force works out to be attractive. Thus, the net force between the two charges (electric and magnetic together) is less than the magnitude of the force they would exert on each other if they were at rest. [ASIDE: This can also be thought of in terms of the transformation properties of forces between different reference frames in special relativity, if you prefer to think of it that way.]
Now, the conservation of electric charge can be written in terms of the continuity equation: $$ \vec{\nabla} \cdot \vec{j}_e + \frac{ \partial \rho_e}{\partial t} = 0 $$ Note that this can be derived from Ampère's Law and Gauss's Law ($\epsilon_0 \vec{\nabla} \cdot \vec{E} = \rho_e$), using the fact that the divergence of a curl is always zero: $$ 0 = \vec{\nabla} \cdot (\vec{\nabla} \times \vec{B}) = \mu_0 \vec{\nabla} \cdot \vec{j}_e + \mu_0 \frac{\partial ( \epsilon_0 \vec{\nabla} \cdot \vec{E})}{\partial t} = \mu_0 \left( \vec{\nabla} \cdot \vec{j}_e + \frac{\partial \rho_e}{\partial t} \right) $$ If we want to extend Maxwell's equations to magnetic charges, we need to have a magnetic version of Gauss's Law and add in a magnetic current term to Faraday's Law: $$ \vec{\nabla} \cdot \vec{B} = \alpha \rho_m \qquad \vec{\nabla} \times \vec{E} = \beta \vec{j}_m - \frac{\partial \vec{B}}{\partial t} , $$ where $\alpha$ and $\beta$ are arbitrary proportionality factors. But if we try to derive a continuity equation for magnetic charge from these two facts (as we did above for electric charge), we get $$ \beta \vec{\nabla} \cdot \vec{j}_m - \alpha \frac{\partial \rho_m}{\partial t} = 0, $$ and this is equivalent to the continuity equation if and only if $\alpha = - \beta$. Beyond this, the choice of $\alpha$ is to some degree arbitrary; different values correspond to different choices of which type of magnetic charge we call "positive", and what units we use to measure it. If we want to have magnetic field lines radiating away from "positive" magnetic charges, then we will want $\alpha > 0$; the usual choice in MKS units is to pick $\alpha = \mu_0$ (and $\beta = -\mu_0$), as you have in your equations above.
This negative sign in the magnetic current term Faraday's Law then implies that the electric field lines created by a moving magnetic charge will obey a "left-hand rule" instead of a "right-hand rule". In other words, the direction of $\vec{E}$ created by a moving magnetic charge would be opposite the direction of $\vec{B}$ created by a moving electric charge. If we still want two magnetic charges moving along parallel tracks to exhibit a lesser force than what they feel when at rest, then we must also flip the sign of the $\vec{v} \times \vec{E}$ term in the Lorentz force law to compensate for this flip.