Here's a long answer leading up to a mathematical measure for the relevance of the Pauli principle. The quantum state of an electron is not just determined by its energy, angular momenta (quantities that lead to good or approximate quantum numbers). Position is also part of the equation. Now, position in QM is not just a simple vector like it is in classical mechanics, it's ruled by a probability distribution. Such a probability distribution might have an exotic form, but it must satisfy a few elementary rules for [physically relevant] probability distributions:
The probability of the particle being inside some given volume is then given by the integral of the probability distribution over that volume. Note that the probability of the particle being at one specific position is zero (some might say it is infinitesimally small).
You can see this is true by considering that the amount $N$ of positions the particle could be in is infinite (if space is dense, which we assume it is) and you're wondering if is in $1$ particular position. The chance of that being true is $1/N = 1/\infty = 0$.
The probability of the particles being anywhere at all must be $1$. We can say with a hundred percent certainty that the particle is somewhere. This is expressed mathematically by requiring the integral of the probability over the entire volume to be $1$.
The probability distribution must be well-behaved. This usually means it must be continuous (can't have any jumps) and must vanish for infinitely large arguments (this is a requirement that comes along with the previous one, since a non-vanishing distribution would not be integrable to $1$ over all space).
Now, in QM the probability distribution is given by the squared modulus of the wavefunction, $|\psi(\vec{r},t)|^2$. If $\psi(\vec{r},t)$ is an exact wavefunction, it contains all the information about the particle(s) it describes. Let's assume it is exact and it describes the state of electron numero uno. Furthermore, let's assume electron numero duo is described by the exact wavefunction $\phi(\vec{r},t)$.
So we surely have two probability distributions linked to this situation. One being $|\psi(\vec{r},t)|^2$, the other $|\phi(\vec{r},t)|^2$. However, there's another probability we might be concerned with: what's the probability that electron numero duo is within a certain volume, given the probability distribution of electron numero uno? This is a conditional probability, which we can calculate as the integral over the product of both wavefunctions where the first one is complex conjugated:
$$\langle\phi(\vec{r},t)|\psi(\vec{r},t)\rangle = \int_{V}{\phi^*(\vec{r},t)\psi(\vec{r},t) d\vec{r}}$$
I've used the conventional notation for this probability in the left hand side of the equation. The right hand side is sometimes called the overlap integral because the integrand is a measure for how strongly both wavefunctions overlap. Note that it becomes the normal probability for a single particle if we take the overlap integral of one wavefunction with itself.
This helps our intuition about the Pauli principle. If the overlap integral is $1$, the electrons must differ in at least one quantum number, e.g. their spin. If it is close to $1$, the electrons should be highly unlikely to be found in the same state. The smaller the overlap integral becomes, the less strictly the Pauli exclusion principle applies. This sounds strange because in its well-known form [no two identical fermions may occupy the same quantum state simultaneously], the Pauli principle looks very strict and law-like. But that's the problem with the well-known form.
Now, we've been assuming the wavefunctions to be exact. This would have been fine if the particles had no way of interacting with each other. However they do. So we need to model that interaction. Electron numero uno influences numero duo and vice versa (little actual English in that sentence). So we'll now treat both electrons as one single system, described by an abstract wavefunction of two spatial arguments $\Psi(\vec{r}_1,\vec{r}_2,t)$.
Let's see what happens if we interchange both electrons. This is sometimes pictured as if we "make them switch places", which makes it seem like we're switching two little balls. Of course this is a bad picture because of the delicate nature of positions in QM we talked about earlier. A better way to think about it is to imagine we changed all the quantum numbers (actually just the entire state except for the probability distributions) of both electrons. So let's do that. We'll denote this switch as a switch of the spatial arguments which suggests the first picture, but this is simply for notational ease.
The operation of interchanging the states is a permutation, we'll call the operator carrying out this task $\hat{P}$. The effect of this operator is clear. We'll drop the time dependence since it is inconsequential here:
$$\hat{P}\Psi(\vec{r}_1,\vec{r}_2) = \Psi(\vec{r}_2,\vec{r}_1).$$
The probability distribution $|\Psi(\vec{r}_2,\vec{r}_1)| = |\Psi(\vec{r}_1,\vec{r}_2)|$ must be unchanged by this operation, so this must mean that
$$\Psi(\vec{r}_2,\vec{r}_1) = \text{e}^{i\delta}\Psi(\vec{r}_1,\vec{r}_2).$$
Furthermore, switching the electrons again should return the wavefunction to its original form, meaning that
$$\text{e}^{2i\delta} = 1$$
or
$$\text{e}^{i\delta} = \pm 1.$$
So this means the act of "switching" the electrons either leaves the wavefunction unchanged, or it yields a sign change. Note that nothing in the above discussion relies on the fact that these are electrons we're describing. So this holds for two general particles. Now, the first option (no sign change upon switching) is associated with bosons, the second one with fermions. So for our electrons a sign change occurs. That's the antisymmetry Oaoa already stressed in his answer. We get a form for the wavefunction out of this property:
$$\Psi(\vec{r}_1,\vec{r}_2) = A\left[\psi(\vec{r}_1)\phi(\vec{r}_2)-\psi(\vec{r}_2)\phi(\vec{r}_1)\right]$$
Note that this wavefunction vanishes if $\psi = \phi$, i.e. if the states of both electrons are the same.
Now using this antisymmetric wavefunction we can find out what the relation is between Pauli's principle (or the antisymmetry of the two-electron wavefunction) and the distance between the electrons. To do this, let's calculate the average squared distance between them. This is given by (let's assume 1D for simplicity of notation)
$$\langle\Psi(x_1,x_2)|(x_1-x_2)^2|\Psi(x_1,x_2)\rangle \equiv \langle(x_1-x_2)^2\rangle = \langle x_1^2\rangle + \langle x_2^2\rangle - 2 \langle x_1x_2\rangle$$
The "unmixed" expectation values can be calculated quite easily and are of course equal. Together they yield
$$\langle x_1^2\rangle + \langle x_2^2\rangle = \langle\psi|x^2|\psi\rangle + \langle\phi|x^2|\phi\rangle.$$
The "mixed" expectation value yields
$$\langle x_1x_2\rangle = \langle\psi|x|\psi\rangle\langle\phi|x|\phi\rangle - |\langle\psi|x|\phi\rangle|^2.$$
Note that the second term in this corresponds to the fermionic permutation $|\psi\rangle \leftrightarrow|\phi\rangle$. These results yield the following expression for the expectation value of the squared distance between the electrons:
$$\langle(x_1-x_2)^2\rangle = \langle\psi|x^2|\psi\rangle + \langle\phi|x^2|\phi\rangle - 2\langle\psi|x|\psi\rangle\langle\phi|x|\phi\rangle + 2|\langle\psi|x|\phi\rangle|^2.$$
What can be learnt from this? Well, for distinguishable particles (say an electron and a muon), which cannot be in the same state, we get the same expectation value except for the last term. This term is called the exchange term. It seems the electrons are on average farther away from each other than, say, an electon and a muon. How much farther is determined by the exchange term. Note that the exchange term is zero if there is no overlap between the wavefunctions (the states are entirely different).
This confirms our earlier intuition, only now we have a mathematical prescription to back it up. We could say that if the average squared distance between the electrons is far greater than the exchange term, the Pauli principle does not contribute much. If the exchange term is comparable to the other terms, the Pauli principle is a lot stricter.
