$T^{ij}$ is nothing more or less than the flow of $i$-momentum across surfaces of constant $j$.1 As a result, the force exerted across a surface $S$ with unit normal one-form components $n_j$ has components
$$ F_{(S)}^i = \int\limits_S T^{ij} n_j \,\mathrm{d}^{d-1}x $$
in $d$ dimensions.
The argument for symmetry is not that the cube is static. The argument is that the cube cannot have infinite angular acceleration as its size shrinks. That is, because we are dealing with a continuous fluid, it should be well behaved as we take our region to be arbitrarily small.
Consider a two-dimensional example of a square covering the area $x_0-L/2 < x < x_0+L/2$, $y_0-L/2 < y < y_0+L/2$. The left surface $x = x_0-L/2$ has $n^\mathrm{left}_j = \delta_j^x$ (sign chosen to correspond to flow of momentum into the square), and so the force on our square due to interactions across the left face has components $F_\mathrm{left}^i \approx L T_\mathrm{left}^{ix}$, where $T_\mathrm{left}^{ix}$ is $T^{ix}$ evaluated at the midpoint $(x_0-L/2,y_0)$. On the opposite surface, $n^\mathrm{right}_j = -\delta_j^x$, so $F_\mathrm{right}^i \approx -L T_\mathrm{right}^{ix}$. Similarly, $F_\mathrm{bottom}^i \approx L T_\mathrm{bottom}^{iy}$ and $F_\mathrm{top}^i \approx -L T_\mathrm{top}^{iy}$.
Torque is a $(d-2)$-form: $\tau = {}^*(\tilde{r} \wedge \tilde{F})$, with $\tilde{r}$ and $\tilde{F}$ the one-forms corresponding to displacement $\vec{r}$ and force $\vec{F}$. In 2D, $\tau = \epsilon_{ij} (r^i F^j - r^j F^i)$. If force components $F_\mathrm{left}^i$ are applied at $(x_0-L/2,y_0)$, then $r_\mathrm{left}^i = -(L/2) \delta^i_x$ and $\tau_\mathrm{left} \approx -(1/2) L^2 T^{yx}$. You can also check $\tau_\mathrm{right} \approx -(1/2) L^2 T^{yx}$ and $\tau_\mathrm{bottom} \approx \tau_\mathrm{top} \approx (1/2) L^2 T^{xy}$.
As we shrink the square down, the midpoints at which we evaluate $T^{ij}$ approach one another and we find the total torque is $\tau \approx L^2 (T_\mathrm{center}^{xy} - T_\mathrm{center}^{yx})$. However, the moment of inertia for a square of surface density $\sigma$ is $\sigma L^4/6$. Thus angular acceleration is
$$ \alpha = \lim_{L\to0} \frac{6(T^{xy}-T^{yx})}{\sigma L^2} $$
at any point in the fluid. Thus we must have $T^{xy} = T^{yx}$, in order to avoid $\alpha \to \infty$. Note that this argument holds in higher dimensions, in more general settings than fluids, and for more general geometries/spacetimes.2
The argument does not however hold in the linear acceleration case. For example, the net $x$-force will have terms like $L (T_\mathrm{left}^{xx} - T_\mathrm{right}^{xx})$ and $L (T_\mathrm{bottom}^{xy} - T_\mathrm{top}^{xy})$. Even though mass is $\sigma L^2$, which would seem to imply linear acceleration goes as $1/L$, the fact is the pairs of stress tensor components naturally cancel as they are evaluated at the same point ($T_\mathrm{left}, T_\mathrm{right}, T_\mathrm{bottom}, T_\mathrm{top} \to T_\mathrm{center}$). No constraints are imposed from this consideration.
1I avoid saying "in the $j$-direction, since we are really interested in surfaces, and these are characterized by one-forms, not vectors. This is more apparent in non-Cartesian (better still non-diagonal) coordinate systems.
2This symmetry always holds, for any material or field, as long as momentum is conserved. You occasionally see reference to the antisymmetric part of the stress tensor, but this comes from splitting the physics into separate domains, and pretending that momentum is lost when going from one to another (e.g. torques can transfer angular momentum from bulk flow into particle spins, and we choose to treat the latter as some momentum-conservation-violating sink as far as the continuum-modeled fluid is concerned).
how on earth.....? A possible way to look at it goes like this, let's consider a little cube length $l$, then the stress force in the $i$th direction acting on $j$th surface element is $-\sigma_{ji}(x_i,x_j,x_k)*dA$ where $dA=l^2$. The force in the $i$th direction acting on the other surface element parallel to the first is $\sigma_{ji}(x_i+dx_i,x_j,x_k)*dA$, so the total stress force acting in the $i$th direction is
$$(\sigma_{ji}(x_i+dx_i,x_j,x_k)-\sigma_{ji}(x_i,x_j,x_k))*dA$$
Now divide by the volume element $dV=dx_idx_jdx_k$ and consider $dA=dx_idx_k$ to obtain
$$f_i^{(j)}=\frac{\sigma_{ji}(x_i+dx_i,x_j,x_k)-\sigma_{ji}(x_i,x_j,x_k)}{dx_j}=\frac{\partial\sigma_{ji}}{\partial x_j}$$
Then total force per unit volume in the $i$th direction is given by your desired equation
Best Answer
Stress is a tensor1 because it describes things happening in two directions simultaneously. You can have an $x$-directed force pushing along an interface of constant $y$; this would be $\sigma_{xy}$. If we assemble all such combinations $\sigma_{ij}$, the collection of them is the stress tensor.
Pressure is part of the stress tensor. The diagonal elements form the pressure. For example, $\sigma_{xx}$ measures how much $x$-force pushes in the $x$-direction. Think of your hand pressing against the wall, i.e. applying pressure.
Given that pressure is one type of stress, we should have a name for the other type (the off-diagonal elements of the tensor), and we do: shear. Both pressure and shear can be internal or external -- actually, I'm not sure I can think of a real distinction between internal and external.
A gas in a box has a pressure (and in fact $\sigma_{xx} = \sigma_{yy} = \sigma_{zz}$, as is often the case), and I suppose this could be called "internal." But you could squeeze the box, applying more pressure from an external source.
Perhaps when people say "pressure is internal" they mean the following. $\sigma$ has some nice properties, including being symmetric and diagonalizable. Diagonalizability means we can transform our coordinates such that all shear vanishes, at least at a point. But we cannot get rid of all pressure by coordinate transformations. In fact, the trace $\sigma_{xx} + \sigma_{yy} + \sigma_{zz}$ is invariant under such transformations, and so we often define the scalar $p$ as $1/3$ this sum, even when the three components are different.
1Now the word "tensor" has a very precise meaning in linear algebra and differential geometry and tensors are very beautiful things when fully understood. But here I'll just use it as a synonym for "matrix."