The fact that the spin of the system is unit means that the system is in the "triplet state." Equally, the net spin of the system is spin 1, you can see following the standard prescription of "addition of angular momentum" for two spin half particles that there are three possible spin states, all of which are symmetric with respect to interchange of the fermions.
$$ \psi_1 = |+\rangle |+\rangle \\
\psi_2 = |-\rangle |-\rangle \\
\psi_3 = |+\rangle |-\rangle + |-\rangle |+\rangle
$$
Fermion statistics require that the net state of an $N$ fermion system be antisymmetric under interchange of any two fermions. Recall that the energy levels of the harmonic oscillator follow
$$ E_n = \hbar \omega \left(n+\frac{1}{2}\right) $$
for the $n$'th harmonic. The specific form of the wavefunction is not important, but you can look them up on wikipedia here. What is important is that any state of the system can be written as
$$ \psi = \sum \limits_i c_i \psi_i(1)\psi_i(2) $$
where $c_i$ is the amplitude for the state and $\psi_i (1,2)$ are the one particle wavefunctions for particle 1 and 2 respectively (written in the "eigenbasis" of some operator with eigenvalues $i$, in this case we take the Hamiltonian so that we have states of definite energy).
We know we can write $\psi$ as the product of the spin part and the spacial part. From above, we can conclude that the spin part of the state is symmetric, therefore, the spacial part must be antisymmetric. The lowest possible energy spacial wavefunction is then (disregarding normalization)
$$
\psi_\text{spacial} = \psi_0(1) \psi_1(2) - \psi_1(1)\psi_0(2)
$$
Here we have set $c_i$ to 0 for all $i$ except this combination since any other combination must have larger values for $n$ or not be antisymmetric. This is the only way to get an antisymmetric combination of $n=0$ and $n=1$, and that's the lowest energy level since trying to write an antisymmetric state with both $n = 0$ would cause a cancellation and result in $\psi = 0$.
Then the total energy of such a state is the sum of the energies of the individual states, namely, $E_\text{net} = E_0 + E_1$, which gives you the answer of $E = 2 \hbar \omega$.
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.
Best Answer
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.