Why, for a spin-½ particle, are the possible outcomes of measuring spin projection along any direction the same? If one measures the projection of spin of a spin half particle along the $x$ axis one will always get $\pm\tfrac12\hbar$.
Measuring it along the $y$ axis one will always get $\pm\tfrac12\hbar$.
Measuring it along the $z$ axis one will always get $\pm\tfrac12\hbar$.
Measuring it along "any" axis one will always get $\pm\tfrac12\hbar$.
Why are the possible outcomes the same in all directions?
 A: The question is "Why are the possible outcomes the same for all directions?"
It happens also for observables of classical physics! QM does not matter here, the truly relevant idea is the fact that in a inertial system physics appears to be isotropic. 
In practice, it is not possible to physically distinguish different directions with physical experiments. 
To illustrate this (definitely non-trivial general) physical phenomenon, suppose to have a physical (classical, quantum, relativistic, quantum-relativistic) system in a certain state $s$.  If you fix a direction, say $x$, and measure some property of the physical system  along that direction, say $A_x$, and then you choose another direction $x'$ with corresponding physical property $A_{x'}$ (obtained by rotating from $x$ to $x'$ the instruments used to measure $A_x$), it is always possible to change the state of the system from $s$ to some $s'$, in order that the outcome of the measurement of $A_{x'}$ is the same as the outcome of the measurement of $A_{x}$.
Changing all possible states and all possible outcomes, you see that the set of values attainable by $A_x$ must be the ones attainable by $A_{x'}$. 
This reasoning can be applied to the components of the momentum, in classical and quantum physics but also to the components of the spin or the a angular momentum. It does not matter if the values are discrete or continuous.  This difference, instead, strongly depends on the used physics, classical or quantum. But this was not your question if I correctly understood.  
ADDENDUM. The value of the spin of an electron along a given direction depends on the state. If the state is prepared as $|z+\rangle$, the outcome of a measurement of $S_z$ is always $+\hbar /2 $ (the analog happens measuring the spin along $-z$), while it is $\pm \hbar /2$ along the other directions with probabilities depending on the state.  However the possible values, changing the state in all possible ways, are the same for all directions: $\pm \hbar/2$
A: Two-valued spin projection is a (quantum) feature of spin-1/2 particles. It is like finding a particle "here" or "there" in a classical physics: it must be somewhere withing only two possibilities, like in two channels. By the way, splitting in two channels have a direct physical meaning, see the experiment of Stern-Gerlach in wikipedia.
A: Although I see already two answers, I’ll add mine.
The situation, as original poster described, is such  for a massive spin-½ particle and is not symmetric in this sense for a massless spin-½ particle (some people say this case is properly named “helicity”, but it is a question of terminology). Why does it matter? Because a massive particle has its center-of-mass frame of reference where it is at rest. In this frame one can apply spatial coordinate rotations and see what will happen to the spin. You’ll see that SO(3) of spatial fixed-point rotations is doubly covered by SU(2), a subgroup of U(2), the natural symmetry group of a two-state quantum system. Some transformations of the spacetime map to transformations of the Hilbert space (ℂ² in this case) of spin states; you can see examples in the earlier answers. There is no essential difference between these x, y, z, or other axes; that’s why you see essentially the same thing each time, although measurements (i.e. observables) are different and non-commuting. Generally, a representation of spacetime groups in the spaces of quantum states is the way how the spin is described at all.
The clause that the particle must be at rest in the reference frame of all these x, y, z is essential. For slow (with respect to the light speed) movements one might not see the difference (and in fact, Pauli’s theory of spin ignored it), but for a rapidly moving particle measuring its spin (in the lab frame) along its velocity is not the same thing as measuring its spin perpendicular to its velocity.
A: It has to do with the act of picking a preferred axis along which to measure the spin projection. Upon observing it, you force the particle into a specific state with a given projection along that (say, the $z$) axis, but now the projection along the other axes are indeterminate, due to the superposition that the particle is now in. They do have a value, it is just uncertain until you actually measure it - but then you screw up the certainty of your first measurement.
What I'm about to state is a bit abusive/misleading, but it may help you visualize things.
Technically speaking, the length squared of the spin vector for a particle of spin $s$ is $\bar{S}^2=s(s+1)$ (I assume $\hbar=1$), which for spin-half particles is $\frac{3}{4}$. The contribution of the $z$ component is $S_z=(\pm\frac{1}{2})^2=\frac{1}{4}$, so the rest must come from the other components. Specifically, since spin-half particles can only have projections of plus or minus half (just like you stated), while the mean value of the other components may vanish, the mean value of their square has to be $\frac{1}{4}$, so together all three components give the total length.
