The conjecture is not true, as some examples show.
Let $D(p)$ denote your number. For primes $p \equiv 1 \textrm{ mod } 4$, we have $D(29)=8$, $D(37)=37$, $D(41)=121$ while $h(29)=h(37)=h(41)=1$.
For primes $p \equiv 3 \textrm{ mod } 4$, we have $D(23)=3$, $D(31)=9$, $D(43)=211$, $D(47)=695$ while $h(23)=h(31)=3$, $h(43)=1$, $h(47)=5$.
The following Pari/GP script calculates $D(p)$:
D(p)=kronecker(-2,p)*matdet(matrix((p-1)/2,(p-1)/2,j,k,cotan(Pi*j*k/p)))/(2^((p-3)/2)*p^((p-5)/4))
(This is a floating-point computation, but it is easy to be rigorous by replacing the entries of the matrix by algebraic numbers). It seems that $D(p)$ grows fast.
EDIT. For primes $p \equiv 1 \textrm{ mod } 4$, divisibility does not hold, for example $h(229)=h(257)=3$ but $D(229) \equiv D(257) \equiv 1 \textrm{ mod } 3$.
For primes $p \equiv 3 \textrm{ mod } 4$, one should be able to relate $D(p)$ with the minus class number $h(\mathbb{Q}(\zeta_p))^-$ as alluded to by user134696 using the analytic class number formula and the fact that $D(p)$ looks like a group determinant on $(\mathbb{Z}/p\mathbb{Z})^\times$, so should be a product over Dirichlet characters mod $p$.
The point is that Dirichlet's class number formula relates $h(p^*)$ with $L(\chi_p,1)$ where $\chi_p$ is the Legendre symbol, but $\chi_p$ is even in the case $p \equiv 1 \textrm{ mod } 4$ so it should have nothing to do with $h(\mathbb{Q}(\zeta_p))^-$ in this case.