Dear Josh, the wave functions are only perfectly symmetric and/or perfectly antisymmetric - each of the factors is - in the case of two particles. As Fabian correctly says, for more than 2 particles, the wave function isn't perfectly symmetric and isn't perfectly antisymmetric with respect to particular transpositions of the two particles. The general character of the wave function's behavior is given by a Young diagram.
That's why what you want to be proved can only be proved for two particles. In the text below, I will therefore assume that there are just two electrons. What you want to be proved is that if the total $L=L_1+L_2$ of the two electrons is even, the wave function is even under the exchange of the two electrons, and it's odd if $L$ is odd.
It's not hard to see. By rotational symmetry, the states with a given value of $L$ form a multiplet with $2L+1$ components because $L_z$ may go from $-L$ to $+L$ with the spacing one. Without the loss of generality, you may focus on the wave function with the maximum value of $L_3$, namely $L_z=L$.
To get this maximum value, you need $L_z=L_{z,1}+L_{z,2}$ to be composed of two equally maximal, equally oriented terms. Because $L_{z,1}$ and $L_{z,2}$ also go between $-L_1$ and $+L_1$, or similarly between $-L_2$ and $+L_2$, it's clear that the only way to get $L_z=L$ is to have $L_{z,1}=L_1$ and $L_{z,2}=L_2$: both $z$-components must be maximum, too. But if it is so, then the wave function is simply
$$ \psi_{L_3=L} = \psi_{L_{z,1}=L_1} \otimes \psi_{L_{z,2}=L_2} $$
For $L_1=L_2$, you can easily see that this wave function has to be symmetric under the permutation of $1$ and $2$. After all, it is the tensor product of the two equal pieces. If you antisymmetrized it, you would get zero. And in fact, the total wave function cannot have any complete symmetry or complete antisymmetry under the exchange of the particles $1,2$ if $L_1\neq L_2$. It's because $L_1^2$ acting on the total wave function gives you $L_1(L_1+1)$ times the wave function, while $L_2^2$ acting on the total wave function gives you $L_2(L_2+1)$. Because the two eigenvalues are not equal for $L_1\neq L_2$, the total wave function can't be symmetric under the exchange of $1,2$.
Again, the question about the symmetry of the orbital wave function only has a sharp answer is there are two particles and if they have the same $L_1=L_2$ - and in that case, the function is symmetric under the permutation.
Similarly, one can prove that the exchange of the two particles with the same half-integer spins $S_1=S_2$ produces a minus sign - assuming that $S_1=S_2$ differs from an integer by $1/2$. To do such things, it's useful to imagine that the components of the multiplet with a given $J$ are organized into a symmetric spintensor.
All the $2J+1$ components of the multiplet with the angular momentum $J$ may be expressed as a completely symmetric "tensor" with 2-valued spinor indices,
$$ T_{abc\dots z} $$
Each index is either $0$ or $1$. The number of ones goes from $0$ to the number of indices $N$ - so there are $N+1$ components of this tensor.
Because each index brings $1/2$ to the total angular momentum, it's clear that the number of indices is $N=2J$. Indeed, then we have $N+1=2J+1$ components.
If there were pairs of indices that are antisymmetric with respect to the exchange of the two indices, one could factorize $\epsilon_{ab}$, a totally antisymmetric object. So only the total symmetry is relevant for our simplest case. Now, the spintensor for an angular momentum $J=J_1+J_2$ object may be written as the symmetrization of
$$ T_{abc\dots z} = T^{1}_{(abc\dots m} T^{2}_{nop\dots z)} $$
where the parentheses represent the complete symmetrization - because the multiplet is represented by the totally symmetric tensor, as we said. The tensors $T^1$ and $T^2$ correspond to the $J_1$ and $J_2$ pieces.
However, if the total angular momentum $J$ differs by an odd number from its maximum value $J_1+J_2$, like in the case of the singlet where $J_1=J_2=1/2$ but $J=0$, then we must factorize those epsilons.
$$ T_{abc\dots z, \,\,{\rm missing\,\,}{mn}} = T^{1}_{(abc\dots l} T^{2}_{op\dots z)} \epsilon^{mn} $$
The epsilon was added to the right hand side to reduce the number of indices by two - i.e. the total spin by one. Because the epsilon is antisymmetric, it changes the symmetry of the whole $T$ under the exchange of the two groups of indices. Each time you reduce the total $J$ by one, the symmetry property changes from symmetry to antisymmetry or vice versa. So the sign obtained from the permutation is given, for $J_1=J_2$, by $(-1)^{J-J_1-J_2}$.
It's also useful to know that the orbital wave functions of a single particle that can be expressed as spherical harmonics $Y_{lm}$ pick the factor of $(-1)^{l}$ under parity i.e. the factor of $(-1)^{l+m}$ under $\theta\to\pi-\theta$.
your calculations are correct, however your beginning assumption is not: the $L=2$ states do not have do be specifically symmetric or antisymmetric. Looking at the representation $1\otimes 1 \otimes 1 =(2\oplus 1 \oplus 0)\otimes 1 = 3\oplus 2\cdot2 \oplus 3\cdot1 \oplus 0$, you can see that you can form $2$ representations of $L=2$. This hints that combined, at each value of $L_z$, the $2$ dimensional space is a standard representation of $S_3$. This is why the two states you calculated in the case $L_z=2$ are not symmetric.
In order to get the correct fully antisymmetric wave function of the three electrons, you need to combine these states with the $s=1$ states of spin (if you're familiar with particle physics, this is a bit how you combine the wave functions of the quarks to get the spin 1/2 baryons like the nucleons).
For the final resolution of your problem, I would recommend using different states though. Setting $|L_{z1}L_{z2}L_{z3}\rangle_o$ the orbital angular momentum states (no ambiguity since $L_1=L_2=L_3=1$) and similarily $|s_{z1}s_{z2}s_{z3}\rangle_s$ for the spin part (no ambiguity since $s_1=s_2=s_3=1/2$), it is usually more simple to consider:
$$
|(12)\rangle_o = \frac{1}{\sqrt 2}(|101\rangle_o-|011\rangle_o)
$$
$$
|(23)\rangle_o = \frac{1}{\sqrt 2}(|110\rangle_o-|101\rangle_o)
$$
$$
|(31)\rangle_o = \frac{1}{\sqrt 2}(|011\rangle_o-|110\rangle_o)
$$
which are linear combinations of the two states you identified in the $L=L_z=2$ subspace. They may not be orthogonal and independent, but they make the symmetrisation process more transparent.
Hope this helps and please tell me if you find some mistakes.
Best Answer
One way to think of the $L$ is that is it related to the eigenvalue of the square of the total angular momentum operator $\hat L\cdot \hat L$. Then $$ \hat L\cdot \hat L \, \psi_{n\ell m}(r,\theta,\phi)=\hbar^2 L(L+1)\psi_{n\ell m}(r,\theta,\phi) $$ Since the length of $\vec L\cdot \vec L$ is necessarily non-negative, the eigenvalue of $\hat L\cdot \hat L$ should be non-negative, which implies for integer $L$ that $L\ge 0$ and thus eliminates the possibility of $L=-1$.
Now $\hat L\cdot \hat L $ commutes with the projection $\hat L_z$ and the states $\psi_{n\ell m}(r,\theta,\phi)$ solutions to the time-independent Schrodinger equation are chosen to have fixed value of $L$. Since $\sqrt{L(L+1)}$ is the "length" of the angular momentum vector, the projection $M_L$ cannot be greater than the length and indeed $L$ by itself is also the largest possible $M_L$ value in a set of eigenstates of $\hat L\cdot \hat L $ with eigenvalue $L(L+1)$.
Finally, if you have three electrons, you need to first combine the first two of them, and then combine the result of this with the last one. Thus, combining three particles with $\ell=1$ will give, in the first step, $L=0,1,2$, and combining these with the last $\ell=1$ will give $$ L_{tot}=(L=0)\times (\ell=1)+ (L=1)\times (\ell=1)+(L=1)\times (\ell=1)= 1+0+1+2+1+2+3\, . $$ Note that the final list of $L_{tot}$ contains $1$ three times and $2$ twice so some values of $L_{tot}$ can be repeated (this never happens with only two angular momenta).
Now the symmetry question. To see that the $L=2$ states are necessarily symmetric start with the $L=2,M_L=2$ state which is necessarily $$ \vert L=2,M_L=2\rangle = \vert \ell_1=1,m_1=1\rangle_1 \vert \ell_2=1,m_2=1\rangle_2\, , \tag{1} $$ and is obviously symmetric w/r to interchange of particle index $1$ and $2$. $\vert L=2,M_L=2\rangle$ must be as (1) (up to an overall phase) because, in the set of states $\vert \ell_1=1,m_1\rangle_1\vert \ell_2=1,m_2\rangle_2 $ there is only one way of constructing an eigenstate of $L_z=L_z^{(1)}+L_z^{(2)}$ with eigenvalue $M=2$, and it is by combining the $m_1=1$ and $m_2=1$ states.
To construct the other states with $L=2$ one should act on $\vert L=2,M_L=2\rangle$ with the other angular momentum operators, which are of the form $L_k= L_k^{(1)}+L_k^{(2)}$. Because $L_k$ is clearly symmetric under permutation of particle index $1$ and $2$, the action of $L_k$ does not change the symmetry properties of states under permutation, so all states with $L=2$ will be symmetric under permutation since the one state $\vert L=2,M=2\rangle$ of (1) is clearly symmetric under permutation.
One can show that the $L=1, M=1$ state is of the form $$ \vert L=1,M=1\rangle=\frac{1}{\sqrt{2}} \left(\vert \ell=1,m_1=1\rangle_1\vert \ell=2,m_2=0\rangle_2 -\vert \ell=1,m_1=0\rangle_1\vert \ell=2,m_2=1\rangle_2\right)\, . \tag{2} $$ One way to see this is by using Clebsch-Gordan tables. Clearly, (2) is antisymmetric. Likewise, one easily shows that the $L=0$ state is fully symmetric. See also this post.