For 3-particle systems the relevant permutation group is $S_3$, which is significantly different from $S_2$ in that not all representations of $S_3$ are 1-dimensional.
Whereas 2-particle states can always be written so they transform back to themselves under permutation of labels, i.e. always possible to write
$$
\psi_a(x_1)\psi_b(x_2)=\frac{1}{2}\psi^+_{ab}(x_1,x_2) + \frac{1}{2}\psi^-_{ab}(x_1,x_2)
$$
with
$$
P_{12}\psi^{\pm}_{ab}(x_1,x_2)=\pm \psi^{\pm}_{ab}(x_1,x_2)
$$
the symmetric and antisymmetric combinations, there are 3-particle states that cannot be made to transform back to a multiple of themselves. Already this is apparent for the triple coupling of spin-1/2 particles: there are two copies of total spin $1/2$ in the decomposition and you cannot arrange the states in $S=1/2$ to be symmetric or antisymmetric under all permutations of particles.
The situation does not improve for 3 spin-1 particles. The fully symmetric states will have total $L=3,1$, the fully antisymmetric states will have $L=0$, but there are states of mixed symmetry with $L=1,2$ (there are two copies of these mixed symmetry states).
Thus, you have no obvious way of knowing if a state with $L=1$ is symmetric or of mixed symmetry.
Of course the spin-statistics theorem does not forbid spin-states of mixed symmetry: the must be combined with spatial stages (also of mixed symmetry) to produce a properly symmetrized total state.
When dealing with angular momentum of a combined system you have two possible basis. Let $\mathscr{H}_1$ be the Hilbert space of particle 1 and $\mathscr{H}_2$ be the Hilbert space of particle 2.
On $\mathscr{H}_1$ you have the basis $|j_1 m_1\rangle$ where $j_1 = 1$ and $m_1 = -1,0,1$. On $\mathscr{H}_2$ you have the basis $|j_2 m_2\rangle$ where $j_2 = 1$ and $m_2 = -1,0,1$. In that sense both $\mathscr{H}_1$ and $\mathscr{H}_2$ are three-dimensional Hilbert spaces.
On the other hand, the combined system is $\mathscr{H}_1\otimes \mathscr{H}_2$. One obvious basis for this space is that associated to the complete set of commuting observables $J_1^2,J_2^2,J_{1z}J_{2z}$. This is the basis
$$|j_1,j_2;m_1,m_2\rangle=|j_1,m_1\rangle\otimes |j_2,m_2\rangle.$$
It is clear that $j_1,j_2=1$ and $m_1,m_2=-1,0,1$. You thus have $9$ basis states. But all basis of a finite dimensional Hilbert space have the same number of element which is its dimension, whatever basis set you use, it will have $9$ states.
For completeness, another natural basis is that of the total angular momentum. You define
$$\mathbf{J}=\mathbf{J}_1\otimes \mathbf{1}+\mathbf{1}\otimes \mathbf{J}_2$$
to get operators $J_z$ and $J^2$. They commute with $J_1^2$ and $J_2^2$ so that you get a basis
$$|j_1,j_2;j,m\rangle$$
This is the basis of total angular momentum. It is a result then, that you can consult for example in Cohen's book Volume 2 on the "Addition of Angular Momentum" chapter, that the possible values for $j$ the eigenvalues of $J^2$ are
$$j=|j_1+j_2|,|j_1+j_2-1|,\dots, |j_1-j_2|$$
Here $j_1=j_2=1$ hence the possible values for $j$ are $$j=2,1,0.$$
Now for $j = 0$ you have just $m = 0$ (one state), for $j = 1$ you have $m = -1,0,1$ (three states) and for $j = 2$ you have $ m = -2,-1,0,1,2$ (five states). This gives a total of $9$ states on the basis of total angular momentum as anticipated.
Best Answer
For every half-integer $j=n/2,n\in\mathbb{Z}$, there is an irreducible representation of $SU(2)$ $$D^j=\exp(-i\vec\theta\cdot\vec J^{(j)})$$ in which the three generators $J_i^{(j)}$ are $(2j+1)\times(2j+1)$ square hermitian matrices. As you probably know, $D^j$ describes states with angular momentum $-j$ to $j$ in integer steps. Given two particles with angular momenta $j$ and $\ell$, we write the total angular momentum of the system as the tensor product $D^j\otimes D^\ell$. We have the standard result $$D^j\otimes D^\ell=\bigoplus_{k=|j-\ell|}^{j+\ell}D^k$$ where on the right it is (conventionally) understood that we write the rotation matrices in order of decreasing angular momentum. Consider, for instance, a meson. This is a composite particle of two spin-half particles. In its ground state, the meson spin representation is $$D^{1/2}\otimes D^{1/2}=D^1\oplus D^0$$ We can thus predict that there are spin-1 and spin-0 mesons, which has been experimentally verified.
Suppose then we wish to find the angular momentum of a baryon. We need $D^{1/2}\otimes D^{1/2}\otimes D^{1/2}$. Convince yourself of the following: Given three matrices $A,B,C$ we have $$A\otimes(B\oplus C)=A\otimes B\oplus A\otimes C$$ Using this, we have $$D^{1/2}\otimes D^{1/2}\otimes D^{1/2}=D^{1/2}\otimes(D^1\oplus D^0)=D^{1/2}\otimes D^1\oplus D^{1/2}\otimes D^0=D^{3/2}\oplus D^{1/2}\oplus D^{1/2}$$ Again, we find baryons with identical quark content in both $3/2$ and $1/2$ states. (There is a slight technicality with the second $D^{1/2}$ involving the Pauli principle.)
The obvious generalization of this is $$D^m\otimes D^j\otimes D^\ell=D^m\otimes\left(\bigoplus_{k=|j-\ell|}^{j+\ell}D^k\right)=\bigoplus_{k=|j-\ell|}^{j+\ell}D^m\otimes D^k=\bigoplus_{k=|j-\ell|}^{j+\ell}\bigoplus_{n=|m-k|}^{m+k}D^n$$ and so on for more particles. From here you can use the Clebsch-Gordan method to construct the actual state vectors for your system.