It is relatively easy to derive the potential energy stored into the magnetic field of an uniformly magnetized sphere of radius $R$ and total magnetic moment $\mu$ : \begin{equation}\tag{1} U_{\text{magn}} = \frac{\mu_0 \, \mu^2}{4 \pi R^3}. \end{equation}
The field is uniform inside the sphere, and dipolar on the exterior side. It is exerting a pressure that I want to calculate. In thermodynamics, the pressure can be defined as the partial derivative of the "internal" energy relative to a change of volume : \begin{equation}\tag{2} P = -\, \frac{\partial U}{\partial V}. \end{equation} Since $V = 4 \pi R^3 / 3$, it is tempting to derive directly (1) (assuming $\mu = \text{constant}$), to get this relation : \begin{align} P_{\text{magn} \, 1} = -\, \frac{\partial \,}{\partial V} \Big( \frac{\mu_0 \, \mu^2}{3 V} \Big) &= \frac{\mu_0 \, \mu^2}{3 V^2} \\[12pt] &\equiv \frac{U_{\text{magn}}}{V}. \tag{3} \end{align} This recalls the stiff equation of state $p = \rho$.
I don't think that $\mu$ could be considered as an independant variable (changing the sphere radius may have an effect on the total dipolar moment, unless there's some kind of constraint). The polar field $B_{\text{pole}}$ is related to $\mu$ by this : \begin{equation}\tag{4} \mu(R, B_{\text{pole}}) = \frac{2 \pi B_{\text{pole}}^2 \, R^3}{\mu_0}. \end{equation} If I assume that $B_{\text{pole}}$ is the independant variable (it may be some constraint), then substituting (4) into (1) and doing the derivative gives this relation instead (a negative pressure = tension !) : \begin{equation}\tag{5} P_{\text{magn} \, 2} = -\, \frac{\partial \,}{\partial V} \Big( \frac{3 B_{\text{pole}}^2}{4 \mu_0} \, V \Big) = -\, \frac{U_{\text{magn}}}{V}. \end{equation} This recalls the cosmological constant relation of state : $p = -\, \rho$.
There's a third possibility (are there others ?). I can consider the magnetic flux $\Phi = B_{\text{inside}} \, \pi R^2$ as the independant variable (flux of the sphere's internal field passing through its own equator) : \begin{equation}\tag{6} \Phi = \frac{\mu_0 \, \mu}{2 R}. \end{equation} In this case, the pressure would be \begin{equation}\tag{7} P_{\text{magn} \, 3} = -\, \frac{\partial \,}{\partial V} \Big( \frac{\Phi^2}{\pi \mu_0 \, R} \Big) = \frac{U_{\text{magn}}}{3 V}, \end{equation} which recalls the electromagnetic equation of state from relativistic physics : $p = \frac{1}{3} \, \rho$.
I suspect that $P_{\text{magn} \, 3}$ should be the proper pressure of the magnetic field. But how to justify this ?
Take note that $U_{\text{magn}}/V$ is NOT the field energy density, since it's varying from one place to another (the field outside the sphere is not uniform, since it's dipolar). So I'm not sure how to interpret the $P$ above correctly, since it's a constant (i.e not depending on position).
So the question is the following :
What is the total magnetic pressure felt by an outside agent that changes a bit the volume of a magnetized sphere ? I expect this : \begin{equation}\tag{8} P_{\text{magn} \, 3} = \frac{U_{\text{magn}}}{3 V} = \frac{\mu_0 \, \mu^2}{(4 \pi R^3)^2} \equiv \frac{B_{\text{int}}^2}{4 \mu_0} \equiv \frac{B_{\text{pole}}^2}{4 \mu_0}. \end{equation}