Argthtjtr

Let $M_n$ be the $ntimes n$ matrix with entries
$$binom{i}{2j}+binom{j}{2i}, qquad text{for $1leq i,jleq n$}.$$

QUESTION. Is this true? There is some evidence. The determinant $det(M_{2n+1})=0$ and
$$det(M_{2n})=(-1)^nbinom{2n}n^22^{n(n-3)}.$$

Noam Elkies in the comments reduces the problem to the identity $$detleft(binom{(n+1)+i}{2j+2}right)_{i,j=0}^{n-1}=binom{2n}n2^{n(n-3)/2}.$$ In general the determinant $binom{N+i}{c_j+j}$, $i,j=0,dots,n-1$ for integers $0leqslant c_0leqslant c_1leqslant dots leqslant c_{n-1}leqslant N$ may be calculated by the following trick.

Consider the Young diagrams $lambda,mu$ with rows length $c_0leqslant c_1leqslant dots leqslant c_{n-1}$ and $N,N,dots,N$ ($n$ rows in total) respectively. Count the skew Young tableaux of the shape $musetminus lambda$. On one hand, the number of such tableaux is known (see Okounkov, A., Olshanski, G.: Shifted Schur functions. St. Petersburg Math. J. 9(2), 73–146 (1997), Theorem 8.1) to be equal to $$frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N+i)!}cdot prod_{i=0}^{n-1} (c_i+i)!cdot detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}.$$
On the other hand, it is the same as the number of usual Young tableaux of the shape $(N-c_0,dots,N-c_{n-1})$, which may be evaluated by Hook Length Formula as
$$
frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
$$
Therefore
$$
detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}=\
prod_{i=0}^{n-1}frac{(N+i)!}{(c_i+i)!(N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).quad (*)
$$
It remains to substitute $N=n+1$ and $c_i=i+2$.
We get $$prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i<j}(j-i)=prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i=0}^{n-1} i!$$
and $(*)$ rewrites as
$$
detleft(binom{(n+1)+i}{2j+2}right)_{i,j=0}^{n-1}=
frac{1!2!dots (n-1)! (n+1)!dots (2n)!}
{2!4!dots (2n)! 2! 4!dots (2n-2)!}2^{n(n-1)/2}.
$$
Replace $2!4!dots (2n)!$ to $$2cdot 1!cdot 4cdot 3!cdot 6cdot 5!cdot ldotscdot (2n)cdot(2n-1)!=\=2^nn! 1!3!dots (2n-1)!.$$ We get $frac{(2n)!}{n!n!} 2^{n(n-3)/2}$
as supposed.

Remark: of course the above determinant is a polynomial in $N$, and if we replace $(N+i)!/(N-c_{n-i-1}+i)!$ in $(*)$ by $c_{n-i-1}! binom{N+i}{c_{n-i-1}}$, the formula becomes true for arbitrary $N$, not necessarily positive integer.

As a follow-up on Fedor's wonderful proof, I would like to pen down just a comment.

Let's replace $c_j+j$ by $u_j$ and treat $N$ as an indeterminate. Then, for non-negative integers $a$ and $b$, we have the determinantal evaluation
$$detleft(binom{N+i+a}{u_{j+b}}right)_{i,j=0}^{n-1}
=prod_{i=0}^{n-1}frac{(N+i+a)!}{(N+n-a-1)!}binom{N+n+a-1}{u_{i+b}}prod_{i<j}^{0,n-1}(u_{j+b}-u_{i+b}).$$
The proof follows from the method of condensation; see Desnanot–Jacobi identity.

Remark. The introduction (or generalization) of the parameters $a$ and $b$ is what makes above-mentioned method work, effectively.