How far can an $N$-fermion wavefunction be from the nearest Slater determinant?

There is no minimum, at least for $N=2$.

Without loss of generality, we can assume that all wavefunctions in the problem have norm $1$.

Restricting to $D=1$, we perform the change of coordinates $u=(x_1+x_2)/sqrt2,quad v=(x_1-x_2)/sqrt2$ and consider the wavefunction

$$psi(u,v)=sqrt{frac{2pi}{varepsilon}} v: e^{-frac{pi}{2}(v^2/varepsilon+u^2varepsilon)}$$

for $varepsilon>0$. This function is even in $u$ and odd in $v$, so it is antisymmetric in the original coordinates. The integral of its square over all space is

$$iint psi(u,v):du:dv= iint frac{2pi}{varepsilon} v: e^{-pi(v^2/varepsilon+u^2varepsilon)} = frac{2pi}{varepsilon} frac{varepsilon^{3/2}}{2pi}varepsilon^{-1/2}=1.$$

On the other hand, we have

$$psi(u,v)= -2 sqrt{frac{2}{pi}}varepsilon: e^{-frac{pi}{2}u^2varepsilon}:frac{d}{dv} left( frac{1}{sqrt{2varepsilon}} e^{-frac{pi}{2}v^2/varepsilon}right).$$

As $varepsilonto 0$, he term inside the derivative is a nascent delta, while the exponential term outside is $1+O(varepsilon)$.

Now we return to the original coordinates, and consider any (normalized) wavefunction of the form

$$s(x_1,x_2)=f(x_1)g(x_2)-f(x_2)g(x_1),$$

as in the problem. The inner product $langle s | psi rangle$ can be computed by using the properties of the Dirac delta derivative to perform the integration on $x_2$. The result is

$$K varepsilon int (f'(x)g(x)-f(x)g'(x)) :dx + O(varepsilon^2),$$

for some unimportant constant $K$. This can be made as small as we want by decreasing $varepsilon$.

If I'm not mistaken, this can be generalized to $D>1$ by taking $psi$ to be the product of the former function in the first pair of coordinates by nascent Dirac deltas in the other coordinates.