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The answer is YES in even dimension. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore $${\rm Pf}(\tilde A)^2=\det(A+iI)\det(A-iI).$$ On the other hand $$\det(A-iI)=\overline{\det(A+iI)}=\det(A+iI)^*=\det(-A-iI)=\det(A+iI).$$ We deduce ${\rm Pf}(\tilde A)^2=\det(A+iI)^2$. Viewing this identityhas a polynomial identity in the entries of $A$, we infer ${\rm Pf}(\tilde A)=\epsilon\det(A+iI)$ for a constant $\epsilon=\pm1$. Taking $A=0_n$, we find $\epsilon=1$.

The odd-dimensional case is interesting too. We write instead $$\det\tilde A=\det(A^2+I)=\det(I+iA)\det(I-iA).$$ With $$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^*=\det(I+iA),$$ we obtain $$\det\tilde A=(\det(I+iA))^2.$$ Finally, we remark that $\det(I+iA)$ is a polynomial in the entries of $A$, with real entries because $I+iA$ is Hermitian. We therefore deduce $${\rm Pf}(\tilde A)=\det(I+iA).$$ For instance, if $n=3$ and the off-diagonal entries of $A$ are $\pm a,\pm b,\pm c$, then $${\rm Pf}(\tilde A)=1-a^2-b^2-c^2.$$ Edit. The above calculation is valid for every dimension, not only the odd one. Remark that ${\rm Pf}(\tilde A)$, as a polynomial in the entries of $A$, is even. In particular, if $n$ is odd, ${\rm Pf}(\tilde A)$ has degree $n-1$ only, whereas if $n$ is even, then ${\rm Pf}(\tilde A)$ has degree $n$.

The answer is YES in every dimension, up to a sign. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore $${\rm Pf}(\tilde A)^2=\det(I+iA)\det(I-iA).$$ On the other hand $$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^*=\det(I+iA),$$ yields $${\rm Pf}(\tilde A)^2=(\det(I+iA))^2.$$ Let us remark that $\det(I+iA)$ is a polynomial in the entries of $A$, with real entries because $I+iA$ is Hermitian. We deduce $${\rm Pf}(\tilde A)=\epsilon\det(I+iA)$$ for some constant $\epsilon=\pm1$. Taking $A=0_n$ gives $\epsilon=1$.

Remark that ${\rm Pf}(\tilde A)$, as a polynomial in the entries of $A$, is even. In particular, if $n$ is odd, ${\rm Pf}(\tilde A)$ has degree $n-1$ only, whereas if $n$ is even, then ${\rm Pf}(\tilde A)$ has degree $n$. For instance, if $n=3$ and the off-diagonal entries of $A$ are $\pm a,\pm b,\pm c$, then $${\rm Pf}(\tilde A)=1-a^2-b^2-c^2.$$

The answer is YES in even dimension. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore $${\rm Pf}(\tilde A)^2=\det(A+iI)\det(A-iI).$$ On the other hand $$\det(A-iI)=\overline{\det(A+iI)}=\det(A+iI)^*=\det(-A-iI)=\det(A+iI).$$ We deduce ${\rm Pf}(\tilde A)^2=\det(A+iI)^2$. Viewing this identityhas a polynomial identity in the entries of $A$, we infer ${\rm Pf}(\tilde A)=\epsilon\det(A+iI)$ for a constant $\epsilon=\pm1$. Taking $A=0_n$, we find $\epsilon=1$.

The odd-dimensional case is interesting too. We write instead $$\det\tilde A=\det(A^2+I)=\det(I+iA)\det(I-iA).$$ With $$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^*=\det(I+iA),$$ we obtain $$\det\tilde A=(\det(I+iA))^2.$$ Finally, we remark that $\det(I+iA)$ is a polynomial in the entries of $A$, with real entries because $I+iA$ is Hermitian. We therefore deduce $${\rm Pf}(\tilde A)=\det(I+iA).$$ For instance, if $n=3$ and the off-diagonal entries of $A$ are $\pm a,\pm b,\pm c$, then $${\rm Pf}(\tilde A)=1-a^2-b^2-c^2.$$ Remark that if $n$ is odd, ${\rm Pf}(\tilde A)$ has degree $n-1$ only, whereas if $n$ is even, then ${\rm Pf}(\tilde A)$ has degree $n$.

The answer is YES in even dimension. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore $${\rm Pf}(\tilde A)^2=\det(A+iI)\det(A-iI).$$ On the other hand $$\det(A-iI)=\overline{\det(A+iI)}=\det(A+iI)^*=\det(-A-iI)=\det(A+iI).$$ We deduce ${\rm Pf}(\tilde A)^2=\det(A+iI)^2$. Viewing this identityhas a polynomial identity in the entries of $A$, we infer ${\rm Pf}(\tilde A)=\epsilon\det(A+iI)$ for a constant $\epsilon=\pm1$. Taking $A=0_n$, we find $\epsilon=1$.

The odd-dimensional case is interesting too. We write instead $$\det\tilde A=\det(A^2+I)=\det(I+iA)\det(I-iA).$$ With $$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^*=\det(I+iA),$$ we obtain $$\det\tilde A=(\det(I+iA))^2.$$ Finally, we remark that $\det(I+iA)$ is a polynomial in the entries of $A$, with real entries because $I+iA$ is Hermitian. We therefore deduce $${\rm Pf}(\tilde A)=\det(I+iA).$$ For instance, if $n=3$ and the off-diagonal entries of $A$ are $\pm a,\pm b,\pm c$, then $${\rm Pf}(\tilde A)=1-a^2-b^2-c^2.$$ Edit. The above calculation is valid for every dimension, not only the odd one. Remark that ${\rm Pf}(\tilde A)$, as a polynomial in the entries of $A$, is even. In particular, if $n$ is odd, ${\rm Pf}(\tilde A)$ has degree $n-1$ only, whereas if $n$ is even, then ${\rm Pf}(\tilde A)$ has degree $n$.

The answer is YES in even dimension. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore $${\rm Pf}(\tilde A)^2=\det(A+iI)\det(A-iI).$$ On the other hand $$\det(A-iI)=\overline{\det(A+iI)}=\det(A+iI)^*=\det(-A-iI)=\det(A+iI).$$ We deduce ${\rm Pf}(\tilde A)^2=\det(A+iI)^2$. Viewing this identityhas a polynomial identity in the entries of $A$, we infer ${\rm Pf}(\tilde A)=\epsilon\det(A+iI)$ for a constant $\epsilon=\pm1$. Taking $A=0_n$, we find $\epsilon=1$.

The answer is YES in even dimension. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore $${\rm Pf}(\tilde A)^2=\det(A+iI)\det(A-iI).$$ On the other hand $$\det(A-iI)=\overline{\det(A+iI)}=\det(A+iI)^*=\det(-A-iI)=\det(A+iI).$$ We deduce ${\rm Pf}(\tilde A)^2=\det(A+iI)^2$. Viewing this identityhas a polynomial identity in the entries of $A$, we infer ${\rm Pf}(\tilde A)=\epsilon\det(A+iI)$ for a constant $\epsilon=\pm1$. Taking $A=0_n$, we find $\epsilon=1$.

The odd-dimensional case is interesting too. We write instead $$\det\tilde A=\det(A^2+I)=\det(I+iA)\det(I-iA).$$ With $$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^*=\det(I+iA),$$ we obtain $$\det\tilde A=(\det(I+iA))^2.$$ Finally, we remark that $\det(I+iA)$ is a polynomial in the entries of $A$, with real entries because $I+iA$ is Hermitian. We therefore deduce $${\rm Pf}(\tilde A)=\det(I+iA).$$ For instance, if $n=3$ and the off-diagonal entries of $A$ are $\pm a,\pm b,\pm c$, then $${\rm Pf}(\tilde A)=1-a^2-b^2-c^2.$$ Remark that if $n$ is odd, ${\rm Pf}(\tilde A)$ has degree $n-1$ only, whereas if $n$ is even, then ${\rm Pf}(\tilde A)$ has degree $n$.