Assume that $A\in M_{nk,nk}$. Then let $(P_n)$ be the sequences of matrices defined as follows: $P_1=J_n,P_2={J_n}^2-I,P_q=J_nP_{q-1}-P_{q-2}$.
Then $\det(A)=\det(P_k)$.
EDIT. The eigenvalues of $J_n$ are the $(1+2\cos(\pi \dfrac{p}{n+1})),p=1\cdots n$; thus, if $Q$ is any polynomial, then we can calculate approximations of the eigenvalues of $Q(J_n)$, and then, an approximation of $\det(A)$. When we have enough significant digits, we deduce the true result -because this one is an integer-. Finally, we do not need to calculate any product of matrices. That follows calculates -using Maple- $\det(A)$ when $n=k=100$ (duration of the calculation: $1$"). First step. You choose Digits:=30. You see that the result has $1076$ digits. Second step. You choose Digits:=1150 and that works (see the sequence of zeros (or eventually of $9$) that appears at the end of the development of the obtained decimal number).
restart; with(LinearAlgebra):
k := 100; n := 100; d := time(); B := Matrix(n, n); for i to n do B[i, i] := 1 end do; for i to n-1 do B[i, i+1] := 1; B[i+1, i] := 1 end do;
z := CharacteristicPolynomial(B, x); u := x; v := x^2-1; for i from 3 to k do w := v; v := rem(vx-u, z, x); u := w end do; Digits := 1150;
Q := unapply(v, x); r := 1; for i to n do r := evalf(rQ(1+2*cos(i*Pi/(n+1)))) end do; print(r); time()-d;
$2.44387846087090290145607732170537391377490420405227812708050615\\
28277319341932844677382952399933460059814926416716644013099963\\
42708968356667589737763656457680692376518632271970928028188495\\
28837548232087652163820090152818313133799717624970641029956038\\
21298982012250961831581518578716473316074214193004344884914447\\
80091522565037919891219503197811771573350002012798682732589728\\
91073456252754229360553614557394171698663316722024355474750138\\
99058808405660398400447542745412413310559180910765198835081950\\
16753460456828320406836683930343030087726159407318434195928328\\
91168720495008933297278988838511004283717390785348840943983494\\
94573265138514209244141811048121198105502888315873129747553394\\
28745956498781145738030840450505861505489488623215771119102138\\
24860932438498432031584839888927118146735452787049842756602723\\
13071431049603803135820994521439844817046772204723218141987299\\
65625418270767015593634878034477052797174424114584736827230518\\
99846006803088990947026408309411889789175194098825709435984858\\
82242334251648224773936990898407542151092941240200527201190067\\ 27465966535881736354100000000000000000000000000000000143203220\\
133487059755244617410791395080\times 10^{1075}$
Choose any matrix with rank $n-1$ that does not have any of the standard unit vectors in its column space.
Added in response to the comment by alex.jordan.
Let $A$ be an $n \times n$ matrix with $rank(A) = n-1$ such that there are vectors $a, \, \tilde a$ with $Aa = 0, \tilde a^T A = 0$ that have all entries $ a_i , \tilde a_i \ne 0$. Then any matrix $B$ that differs from $A$ in exactly one entry has full rank, i.e. $\det B \ne 0$.
To prove this, consider such a $B$. After permuting rows and columns and rescaling, we may assume that $B_{1,1} = A_{1,1} + 1$.
First note that the first column of $A$ is a linear combination of columns $2, \dots, n$, since $Aa = 0$ and $a_1 \ne 0$. The column space of $B$ certainly contains the column space of $A$ and thus $rank(B) \ge n-1$.
If $rank(B) < n$, then $A$ and $B$ must therefore have the same column space. Hence the standard unit vector $e_1$ is in the column space of $A$, that is $Ac = e_1$ for some vector $c$. But then $0 = \tilde a^TAc = \tilde a e_1 = \tilde a_1$, contradicting the assumptions for the left and right null vectors of $A$.
Therefore $rank(B) > n-1$ and $\det B \ne 0$.
Best Answer
There is such a thing, at least over the reals. Suppose $m>n$. Then an $m\times n$ matrix has full rank if and only if it contains an $n\times n$ submatrix of full rank. Let $A$ be an $m\times n$ matrix and let $A_1,\dots,A_N$ be its $n\times n$ submatrices. (The exact value of the number $N$ is irrelevant here; it only depends on $m$ and $n$.) Now let $D(A)=\sum_{k=1}^N\det(A_k)^2$. Clearly $D(A)$ is polynomial in each element since the determinant is, and $D(A)=0$ if and only if none of the $n\times n$ submatrices of $A$ has full rank.
I don't know if such things have been studied or given a name.