Let $H_n$ be an $n×n$ Hadamard matrix and $R_n$ the $n×n$ reverse identity matrix.

The matrix $X= \begin{pmatrix}
H_n & R_nH_n \\
H_n & -R_nH_n
\end{pmatrix}$ has entries of length $1$ and $$XX^* = 2nI_{2n} + ((nI_n - R_nH_nH_n^*R_n) \otimes R_2)$$ which is simply $2nI_{2n}$ so it is a Hadamard matrix. Permute the last $n$ columns with $R_n$ and you have it in the form you give. This gives half-skew-centrosymmetric Hadamard matrices of twice the size of a Hadamard matrix. (It also works for complex Hadamard matrices or even Hadamard matrices over *-rings)

Another construction (edit): suppose we apply the [Paley construction][1] II to a finite field with $q=4k+1$ elements giving an $2n×2n$ Hadamard matrix $H= \begin{pmatrix}
H_1 & H_2 \\
H_3 & H_4
\end{pmatrix} $ after permuting the odd columns and rows to the first $n$ rows/columns, the even as last $n$ AND put the first even row/column at the last position. Define $X := X_{i,j} = (i-j)^{0.5(q-1)}$ (in the finite field this is $\pm 1$ with zero diagonal and symmetric because $-1$ is a square). Note that $X_{i,j+k}=X_{j+k,i}=X_{n-i+k,n-j}$. Then for Paley construction II, using $j$ as all-ones vector $$H_2 = H_3^T = \begin{pmatrix}
j & X-I \\
-1 & j^T
\end{pmatrix}$$ $$H_1 = \begin{pmatrix}
1 & j^T \\
j & X+I
\end{pmatrix}$$ $$H_4 = \begin{pmatrix}
-X-I & -j \\
-j^T & -1
\end{pmatrix}$$The $H_i$ now have the properties $R_nH_2R_n = H_3$ and $-R_nH_1R_n = H_4$ due to our note above. Thus $H$ is a half-skew-centrosymmetric Hadamard matrix!


  [1]: https://en.m.wikipedia.org/wiki/Paley_construction