You're question is a bit unclear and hopefully I'm not misunderstanding anything.

If all the atom points in the same direction $\hat{\mathbf{n}}$ then your state $$ \vert{\psi}\rangle=\vert\hat{\mathbf{n}}\rangle_1 \otimes \vert\hat{\mathbf{n}}\rangle_2\ldots \vert\hat{\mathbf{n}}\rangle_N $$ is actually a coherent state. There is a unique rotation $R(\hat{\mathbf{n}})$ that will take the spin-up state $\vert +\rangle$ (or the spin-down state $\vert -\rangle$ if you prefer) to $\vert\hat{\mathbf{n}}\rangle$ and then you have $$ \vert{\psi}\rangle=[R(\hat{\mathbf{n}})\vert+ \rangle_1] \otimes [R(\hat{\mathbf{n}})\vert+ \rangle_2]\ldots [R(\hat{\mathbf{n}})\vert+ \rangle_N] $$ and there is no definition of squeezing where the coherent state is squeezed. In fact, the coherent state is used to define the standard quantum limit.

In addition, there are several definitions of squeezing but the only one that makes sense on the sphere is the so-called covariant squeezing. Given any state $\vert \phi\rangle$, you first compute the triple $(\langle S_x\rangle, \langle S_y\rangle,\langle S_z\rangle)$ and you use these expectation values to define the direction $\hat{\mathbf{m}}$ of your quantization axis. Thus, the relevant observables are $\hat S_k^\prime=R(\hat{\mathbf{m}})\hat S_k R^{-1}(\hat{\mathbf{m}})$ and you then measure squeezing by comparing the variances of $\hat S_x^\prime$ and $\hat S_y^\prime$ in the plane perpendicular to the direction $\hat{\mathbf{m}}$ on the sphere with the corresponding variance of $\hat S_y^\prime$ and $\hat S_x^\prime$ for the coherent state. This guarantees you get genuine squeezing rather than just a projection effect.

If you're into Wigner function of the sphere, the Wigner function for the relevant coherent state is a Gaussian-like blob pointing the $\hat{\mathbf{m}}$ direction and squeezing is manifested by a deformation of the blob, in particular by the loss of cylindrical symmetry about the $\hat{\mathbf{m}}$ axis.

Because coherent state are just rotated $\vert SS\rangle$ states (or $\vert S,-S\rangle$, you cannot get squeezing by simple rotation from the North or South pole. In particular, any Hamiltonian linear in the angular momenta $$ \hat H=\sum_k \alpha_k S_k $$ generates are rotation by exponentiation so such Hamiltonians cannot produce squeezing. This is why Kitagawa and Ueda generate their squeezing using Hamiltonians quadratic in the generators.

Now, using the definition of Klimov and Chumakov

Klimov, Andrei B., and Sergei M. Chumakov. A group-theoretical approach to quantum optics: models of atom-field interactions. John Wiley & Sons, 2009.

the Dicke state is NOT a state where all spins point in one direction. Rather it is a symmetrized sum $$ \vert k,A\rangle = \sqrt{\frac{k!(A-k)!}{A!}}\sum_p \vert j_1,1\rangle\ldots j_k,1\rangle \vert j_{k+1},0\rangle\ldots \vert j_A,0\rangle $$ where the sum is over all the possible permutations that interchange excited and non-excited atoms (basically, 0's and 1's).

You're question is a bit unclear and hopefully I'm not misunderstanding anything.

If all the atom points in the same direction $\hat{\mathbf{n}}$ then your state $$ \vert{\psi}\rangle=\vert\hat{\mathbf{n}}\rangle_1 \otimes \vert\hat{\mathbf{n}}\rangle_2\ldots \vert\hat{\mathbf{n}}\rangle_N $$ is actually a coherent state. There is a unique rotation $R(\hat{\mathbf{n}})$ that will take the spin-up state $\vert +\rangle$ (or the spin-down state $\vert -\rangle$ if you prefer) to $\vert\hat{\mathbf{n}}\rangle$ and then you have $$ \vert{\psi}\rangle=[R(\hat{\mathbf{n}})\vert+ \rangle_1] \otimes [R(\hat{\mathbf{n}})\vert+ \rangle_2]\ldots [R(\hat{\mathbf{n}})\vert+ \rangle_N]:= R(\hat{\mathbf{n}})\vert SS\rangle $$ where $S=N/2$ and there is no definition of squeezing where the coherent state is squeezed. In fact, the coherent state is used to define the standard quantum limit.

In addition, there are several definitions of squeezing but the only one that makes sense on the sphere is the so-called covariant squeezing. Given any state $\vert \phi\rangle$, you first compute the triple $(\langle S_x\rangle, \langle S_y\rangle,\langle S_z\rangle)$ and you use these expectation values to define the direction $\hat{\mathbf{m}}$ of your quantization axis. Thus, the relevant observables are $\hat S_k^\prime=R(\hat{\mathbf{m}})\hat S_k R^{-1}(\hat{\mathbf{m}})$ and you then measure squeezing by comparing the variances of $\hat S_x^\prime$ and $\hat S_y^\prime$ in the plane perpendicular to the direction $\hat{\mathbf{m}}$ on the sphere with the corresponding variance of $\hat S_y^\prime$ and $\hat S_x^\prime$ for the coherent state. This guarantees you get genuine squeezing rather than just a projection effect.

If you're into Wigner function of the sphere, the Wigner function for the relevant coherent state is a Gaussian-like blob pointing the $\hat{\mathbf{m}}$ direction and squeezing is manifested by a deformation of the blob, in particular by the loss of cylindrical symmetry about the $\hat{\mathbf{m}}$ axis.

Because coherent state are just rotated $\vert SS\rangle$ states (or $\vert S,-S\rangle$, you cannot get squeezing by simple rotation from the North or South pole. In particular, any Hamiltonian linear in the angular momenta $$ \hat H=\sum_k \alpha_k S_k $$ generates are rotation by exponentiation so such Hamiltonians cannot produce squeezing. This is why Kitagawa and Ueda generate their squeezing using Hamiltonians quadratic in the generators.

Now, using the definition of Klimov and Chumakov

Klimov, Andrei B., and Sergei M. Chumakov. A group-theoretical approach to quantum optics: models of atom-field interactions. John Wiley & Sons, 2009.

the Dicke state is NOT a state where all spins point in one direction. Rather it is a symmetrized sum $$ \vert k,A\rangle = \sqrt{\frac{k!(A-k)!}{A!}}\sum_p \vert j_1,1\rangle\ldots j_k,1\rangle \vert j_{k+1},0\rangle\ldots \vert j_A,0\rangle $$ where the sum is over all the possible permutations that interchange excited and non-excited atoms (basically, 0's and 1's).