Interpretation of equal absolute values of pressure and tension in the electric or magnetic field along Cartesian axes aligned with the field For an electric or magnetic field along the $x$ axis, the stress-energy tensor in mixed covariant-contravariant form, in $(t,x,y,z)$ coordinates, is of the form $\operatorname{diag}(1,1,-1,-1)$ (ignoring an over-all constant, in +--- signature). (This has a trace of 0, which Cham explains in comments is a particular feature of 3+1 dimensions.)
At a more elementary level, this tells us that field has tension parallel to a field line, and pressure along the two axes perpendicular to it, and that these three numbers are all equal in absolute value. In other words,
$$|\text{tension in the $x$ direction}|=|\text{pressure in the $y$ direction}|=|\text{pressure in the $z$ direction}|.$$
The second equation is obviously required by symmetry, but the first seems more mysterious to me.
Is there any simple interpretation of this, suitable for freshman physics? Is it an accident of the fact that we live in 3+1 dimensions, or is it not an accident, so that in $n$+1 dimensions we would still have something of the form $(n-2,1,-1,\ldots,-1)$?
 A: First thing to note is the equation 
$$ \text{energy density} = |\text{tension along the $x$ direction}| $$
(in units with $c=1$) would be valid for any dimension, not just $4$. The reason is simple: for purely magnetic (or electric) field along the $x$ axis, the field strength tensor is invariant under boosts along the direction of the field. So the stress energy tensor for two reference frames connected by such a boost must also be the same. But this could only happen if projection of this tensor on $(t,x)$ plane is proportional to the induced metric on this plane. This argument does not depend on the dimensionality space orthogonal to the field direction and is also easily generalized to $p$-form field electrodynamics.
From Minkowski spacetime perspective a non-null electromagnetic field (a field which has non-zero  invariants) at a point defines a timelike plane in a tangential space that we could interpret as tangential to the world volume of a fluxline. There is no preferred direction inside this plane and no preferred direction orthogonal to it and thus the overall structure of stress-energy tensor in 4D at this point is determined by energy density and the tracelessness (in 4D only) of the stress-energy tensor. In higher dimensions we would need additional arguments to fix proportionality constant between tension in the pressure.
But equality of tension and pressure magnitude could also be established by calculating fluxtube energy. Let us consider a small fluxtube element, carrying the flux $\Phi$ (for simplicity let us assume it is magnetic flux only), with length $ l$ and section area $S$. Assuming the field strength in it is almost constant, the energy of this fluxtube element is:
$$
 E = \frac{1}{2\mu_0} \left(\frac{\Phi}{ S}\right)^2  S  l =
\frac{1}{2\mu_0} \frac{\Phi^2  l} { S}\,.
$$
The energy of the fluxtube element is proportional to its length and inversely proportional to its section area. And that is the reason magnitudes of tension and pressure are the same. Indeed, the flux is constant during the evolution of fluxtube, and varying its geometry we would obtain expressions for tension along the field line:
$$
\text{|tension|}=\left(\frac{\partial E} {\partial l} \right) \frac1S,
$$
and for pressure orthogonal to field direction:
$$
\text{pressure}= - \left(\frac{\partial E} {\partial S} \right) \frac1l.
$$
But by  Euler's homogeneous function theorem:
$$
E = l \left(\frac{\partial E} {\partial l} \right) =  - S \left(\frac{\partial E} {\partial S} \right),
$$
so the pressure and tension have equal absolute values. 

If calculations with an infinitesimal fluxtube element are not intuitive enough for freshman physics, one could instead consider a finite but highly symmetric field configuration that could provide similar derivation of tension/pressure. In particular, it might be more intuitive to reason about higher dimensional Coulomb's law than about higher dimensional fluxes.
For example, let us  consider a (hyper)spherical capacitor in $d+1$-dimensional spacetime consisting of two concentric spheres with radii $R$ and $R+l$ (with separation between spheres $l \ll R$) carrying charges $Q$ and $-Q$. Radii of both spheres could be changed independently, while their charges are held fixed. The $d$-dimensional spatial volume filled with electrostatic field is $$V_d=S_{d-1} R ^{d-1} l\,,$$ where $S_n$ is the volume of unit $n$-sphere.
The electrostatic energy of the capacitor:
$$ 
 E = \frac{C_d}{2}  \left(\frac{Q}{S_{d-1}R^{d-1}}\right)^2 V_d = \frac{C_d}{2}   \frac{Q^2 l}{S_{d-1} R^{d-1}}\,,
$$
where the constant $C_d$ depends on the conventions used for higher dimensional electrodynamics. 
Energy density:
$$ \epsilon = \frac{ E}{V_d}. $$
Varying $l$ while keeping $R$ fixed we obtain tension component $\sigma_{rr}$:
$$\text{|tension|} = \left(\frac{\partial E}{\partial V_d}\right)_{R\text{ fixed}}=\frac{\partial E}{\partial l}\left(\frac{\partial V_d}{\partial l}\right)^{-1} = \epsilon,
 $$ 
while varying $R$, keeping $l$ fixed we obtain pressure components $\sigma_{\theta_i\theta_i}$ for hyperspherical coordinates $\{\theta_i\}$:
$$
\text{pressure} = -\left(\frac{\partial E}{\partial V_d}\right)_{l\text{ fixed}}=-\frac{\partial E}{\partial R}\left(\frac{\partial V_d}{\partial R}\right)^{-1} = \epsilon,
$$
So the stress-energy tensor in $d+1$–dimensional spacetime is $T^{\mu}_\nu = \epsilon\, \mathop{\mathrm{diag}}(1,1,-1,-1,\ldots)$. Obviously, its trace is nonzero in dimensions other than $4$.
A: Here is an elementary argument for 3+1 dimensions, which I'll then generalize to $d+1$ dimensions, although I'm unsure if I got the latter right, and would appreciate comments. In the case of a purely electric (or purely magnetic) field, we expect the pressure and tension to depend only on the field at that point -- not on the field in some neighborhood, or on derivatives of the field. Furthermore, we know on dimensional grounds that the pressure and tension must be proportional to the square of the field. Therefore, we only need one concrete example in which we know the answer in order to fix the constant of proportionality for the pressure $a_p$, and likewise for the tension $a_t$.
Consider point charges $q$ and $-q$ at rest. Their attraction must be proportional to the integral of the tension over the plane that is equidistant from the two charges. Since the tension is proportional to the square of the field, this gives an integral $\int\cos^2\theta r^{-4} dS$, which with the omission of some constant factors represented by $(\ldots)$ can be simplified to
$$\frac{1}{a_t}\left|\int T\ dS\right| = (\ldots) \int_0^{\pi/2} \cos^3\theta\sin\theta \ d\theta.$$
In the case of like charges, the integral is $\int\sin^2\theta r^{-4} dS$, which with the omission of the same constant factors equals
$$\frac{1}{a_p}\left|\int P\ dS\right| = (\ldots) \int_0^{\pi/2} \sin^3\theta\cos\theta \ d\theta.$$
These two definite integrals are equal, and since the forces in these two cases are equal, we have $a_t=a_p$.
Generalizing to $d$+1 dimensions, the fields fall off like $r^{-d+1}$ by Gauss's law. We want to make a surface that separates the two charges, so that any flow of momentum from one charge to the other has to pass through the surface. This requires a $d-1$-dimensional surface, so that $dS\sim r^{d-2}dr$. (This is the part of this argument that I'm unsure about.) The integrals then become
$$\int_0^{\pi/2} \cos^d\theta\sin\theta \ d\theta = \frac{1}{d+1}.$$
and
$$\int_0^{\pi/2} \sin^3\theta\cos^{d-2}\theta \ d\theta = \frac{2}{d^2-1},$$
which are unequal for $d\ne3$.
So if I've got this right, then the equality of pressure and tension is an accident of the fact that we live in three dimensions.
A: I'm considering a local inertial frame, at a point $\mathcal{P}$ of $D$ dimensional spacetime, in which there's an electric field (any configuration).  The calculations below are valid even in General Relativity, and for any electric field in spacetime of $D$ dimensions (I don't care about axes aligned to the field!).  I'm using $c = 1$ and metric signature $(1, -1, -1, \dots, -1)$.  In the local frame, there is no magnetic field:
\begin{align}\tag{1}
F_{0 i} &\ne 0, &F_{ij} &= 0.
\end{align}
We can write $F_{0i} = E \, n^i$ (and $F_{i0} = -\, E \, n^i$) where $n_i$ are the components of some unit vector in the $D-1$ dimensional space.  The local frame $D$-velocity has components $u^a = (1, 0, 0, \dots, 0)$, and the local electric field line has an orientation represented by a spacelike $D$-vector: $n^a = (0, n^1, n^2, \dots, n^{D-1})$.  The orthogonal directions (the hyperplane perpendicular to the field line) are represented by $D - 2$ spacelike $D$-vectors $w^a = (0, w^1, w^2, \dots w^{D-1})$.  These unit $D$-vectors are such that
\begin{align}
u_a \, u^a &= 1, &n_a \, n^a &= -1, &w_a \, w^a &= -1, \tag{2} \\[12pt]
u_a \, n^a &= 0, &u_a \, w^a &= 0, &n_a \, w^a &= 0. \tag{3}
\end{align}
We then have the following identities, valid only because there is no magnetic field (write an antisymetric matrix for $F_{ab}$ and these would become obvious):
\begin{align}\tag{4}
F_{ab} \, u^b &= E \, n_a, &F_{ab} \, n^b &= E \, u_a, &F_{ab} \, w^a &= 0.
\end{align}
The general energy-momentum of the field has the following components in $D$ dimensional spacetime:
\begin{equation}\tag{5}
T_{ab} = F_{ac} \, F^c_{\;\, b} + \frac{1}{4} \, \eta_{ab} \, F_{cd} \, F^{cd}.
\end{equation}
Here:
\begin{equation}\tag{6}
\frac{1}{4} \, F_{cd} \, F^{cd} = \frac{1}{2} \, F_{0i} \, F^{0i} = -\, \frac{1}{2} \, F_{0i} \, F_{0i} = -\, \frac{1}{2} \, E^2 \equiv -\, \rho.
\end{equation}
Okay, we have everything.  Contract (5) on the $D$-vectors introduced above:
Energy density of the local field:
\begin{align}
T_{ab} \,u^a \, u^b
 &= -\, (F_{ca} \, u^a)(F^c_{\;\, b} \, u^b) + \frac{1}{4} (\eta_{ab} \, u^a \, u^b) \, F_{cd} \, F^{cd} \\
&= -\, (E \, n_c)(E \, n^c) + \frac{1}{4} \, F_{cd} \, F^{cd} = +\, E^2 - \rho = 2 \rho - \rho = \rho. \tag{7}
\end{align}
Parallel Pressure along the local field line:
\begin{align}
p_{\parallel} \equiv T_{ab} \, n^a \, n^b
 &= -\, (F_{ca} \, n^a)(F^c_{\;\, b} \, n^b) + \frac{1}{4} (\eta_{ab} \, n^a \, n^b) \, F_{cd} \, F^{cd} \\
&= -\, (E \, u_c)(E \, u^c) - \frac{1}{4} \, F_{cd} \, F^{cd} = -\, E^2 + \rho = -\, 2 \rho + \rho = -\, \rho. \tag{8}
\end{align}
Thus $p_{\parallel} = -\, \rho$.  The field line is under tension.
Orthogonal pressure, perpendicular to the local field line:
\begin{align}
p_{\perp}(w) \equiv T_{ab} \, w^a \, w^b
 &= -\, (F_{ca} \, w^a)(F^c_{\;\, b} \, w^b) + \frac{1}{4} (\eta_{ab} \, w^a \, w^b) \, F_{cd} \, F^{cd} \\
&= -\, (0)(0) - \frac{1}{4} \, F_{cd} \, F^{cd} = \rho. \tag{9}
\end{align}
Thus $p_{\perp} = \rho$.  The field lines are pushing against themselves in the orthogonal hyperplane.
These results are what you wanted to show.  The extension to the pure magnetic field is really non-trivial when $D \ne 4$, because the magnetic field cannot be representend by a vector in general (it's a scalar field in $D = 3$ for example.  There is no magnetic field for $D = 2$.  In the case $D=5$, the electric field has 4 components while the magnetic field has 6 components!).  Since $F_{ij}$ is antisymetric, its eigen-vectors have complex components.  It is easy to get $T_{ab} \, u^a \, u^b = \rho$ but the rest is much more complicated and I didn't verified that the results are the same.

EDIT
Now I have a question:  could we say there's a magnetic field "line", in any spacetime of $D \ne 4$ dimensions?  For example, in $D = 3$ dimensions spacetimes, the magnetic field is a scalar field in 2D space.  How can there be a field line in this case?  In $D = 5$ dimensions spacetimes, the electric field has 4 components and is still a vector in 4D space (so the electric field line still have a physical sense).  The magnetic field have 6 components and thus isn't a vector in 4D space.  How can there be a field line in this case?  So we cannot speak of tension "along the field line" and pressure in the "orthogonal hyperplane", isn't?
