The Jacobi identity for the Poisson bracket does indeed follow from the fact that $d\Omega =0$.
I claim that (twice) the Jacobi identity for functions $f,g,h$ is precisely $$d\Omega(X_f,X_g,X_h) = 0.$$
To see this, simply expand $d\Omega$.
You will find six terms of two kinds:
three terms of the form $$X_f \Omega(X_g,X_h) = X_f \lbrace g,h \rbrace$$
and three terms of the form $$\Omega([X_f,X_g],X_h).$$
To deal with the first kind of terms, notice that from the definition of $X_f$, for any function $g$, $$X_f g = \lbrace g, f \rbrace.$$ This means that $$X_f \Omega(X_g,X_h) = \lbrace \lbrace g,h \rbrace, f \rbrace.$$
To deal with the second kind of terms, notice that $$\iota_{[X_f,X_g]}\Omega = [L_{X_f},\iota_{X_g}]\Omega,$$ but since $d\Omega=0$, $$L_{X_f}\Omega = d \iota_{X_f}\Omega = 0,$$ and hence $$\iota_{[X_f,X_g]}\Omega = d \iota_{X_f}\iota_{X_g}\Omega = d\lbrace g,f\rbrace,$$ whence $$\Omega([X_f,X_g],X_h) = d\lbrace g,f\rbrace (X_h) = \lbrace\lbrace g,f\rbrace h \rbrace.$$$$\Omega([X_f,X_g],X_h) = d\lbrace g,f\rbrace (X_h) = \lbrace\lbrace g,f\rbrace, h \rbrace.$$
Adding it all up you get twice the Jacobi identity.