Crossing symmetry is basically the CPT theorem applied in the context of the LSZ formula, using microcausality to re-order the field operators. The role of the CPT theorem is to relate particles to their antiparticles. The CPT transform is not unique: in particular, we can compose it with any proper Poincaré transform to get another equally-good CPT transform. The CPT theorem defines a conjugate relation between two *sets* of single-particle states (like between the set of *all* single-electron states and the set of *all* single-positron states), but it doesn't define a unique one-to-one relationship between individual single-particle states. So questions like "does the antiparticle of a spin-up electron have spin up or spin down?" don't have unique answers. The answer is convention-dependent. 

Crossing symmetry doesn't require choosing a specific CPT transform, but since the relationship between individual single-particle and single-antiparticle states is convention-dependent, we can compensate for a minus sign that comes from Fermi statistics by switching conventions, as Peskin & Schroeder wrote.

Crossing symmetry is not an "ordinary symmetry" that relates physical states to other physical states, and maybe this limits its utility as Weinberg suggested, but neither of these points contradict what Peskin & Schroeder wrote.

For perspective, consider a time-ordered correlation function
$$
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
\newcommand{\dpsi}{\psi^\dagger}
	\la 0|T\,X_A(x) \psi_a(y)|0\ra
$$
where $\psi_a(y)$ is an individual field operator with Lorentz index $a$ and at the spacetime point $y$, and where $X_A(x)$ is an abbreviation for some product of field operators with indices collectively denoted $A$ and spacetime points collectively denoted $x$. If $\psi$ is a fermion field, then the overall sign of the correlation function (and hence of the scattering amplitude) is affected by how the field-operator factors are ordered.

Starting with this correlation function, we can use the LSZ reduction formula to construct a scattering amplitude in which the particle associated with $\psi$ is either in the initial state or in the final state. CPT says that the single-particle part of the state $\psi_a(y)|0\ra$ is an antiparticle of the single-particle part of the state $\la 0|\psi_a(y)$, or equivalently of the state $\dpsi_a(y)|0\ra$. The idea behind LSZ is that we can isolate the desired single-particle contributions to the in/out states by isolating the associated poles. The field operator $\psi_a$ can be written as the sum of its positive- and negative-frequency parts, $\psi_a(y)=\psi_a^+(y)+\psi_a^-(y)$, which act on a state-vector (ket) to their right as annihilation and creation operators, respectively, and conversely when acting on a state-vecctor (bra) to their left. The LSZ formula uses this to select one of the two poles, either incoming or outgoing. The identitities
$$
	\big(\psi_a^+\big)^\dagger
=
	\big(\dpsi_a\big)^-
\hskip2cm
	\big(\psi_a^-\big)^\dagger
=
	\big(\dpsi_a\big)^+
$$
say that the particles corresponding to these two poles are antiparticles of each other. Crossing symmetry amounts to a relationship between the formulas that LSZ uses to select either of these two poles. So in general, what crossing symmetry does to the crossed particle's spin-state is determined by the relationship between the single-particle parts of the states $\psi_a|0\ra$ and $\la 0|\psi_a$. We don't need the LSZ context for this, and we don't need to choose a specific convention for this, either.

> is crossing symmetry trivial for spin-1 particles like it is for scalars?

Crossing symmetry for spin-1 particles (like photons) doesn't have any minus signs from Fermi statistics, but the amplitudes still involve specific components of the field operators (as in the spin-1/2 case), which can be handled according to the general principle described above. Equations (13.5.1)-(13.5.9) in Weinberg give a photon example.

Section 2.1 in https://arxiv.org/abs/1605.06111 gives some convention-dependent details for the case of a "vector, Dirac, and left- or right-handed (massless) Weyl representation respectively" with a footnote that says "the overall sign that relates $u^\sigma$ with $v^{-\sigma}$ ... is conventional since it depends on the choice of the CPT phase." Section 3 in the same paper shows some detailed examples for various spins, for both massive and massless particles.