Why not go one step further and make all the electrons have spin down? Then you should get $S = -3/2$, no?
In fact, what's happening is: you are confusing the projection of the spin quantum number, $M_S$, with its magnitude, represented by the spin quantum number $S$.
The $\pm1/2$ that you are summing are the individual projections of the electron spins, $m_s$. When you add up projections, you get a projection $M_S$. This projection can indeed be $-1/2$, and it can indeed be $-3/2$ as well, but the term symbol does not depend on $M_S$; it depends on $S$.
To obtain the total spin quantum number $S$ you have to use the Clebsch-Gordan series to couple different sources of angular momenta together. In this case we would first couple the spins of two electrons
$$S_{12} = s_1 + s_2, s_1 + s_2 - 1, \cdots, |s_1 - s_2|$$
Note that for an electron, it has spin $s = 1/2$. Again, the projection of the spin can be $+1/2$ or $-1/2$, but that is not what we are interested in. We are interested in the magnitude of its spin, which is only $1/2$, not $-1/2$.1 If we plug $s_1 = s_2 = 1/2$ into that equation above we get
$$S_{12} = 1, 0$$
and now if we couple this to the spin of the third electron we get
$$S = S_{12} + s_3, S_{12} + s_3 - 1, \cdots |S_{12} - s_3|$$
Again, $s_3 = 1/2$ only. But $S_{12}$ can take on two values so we will have to evaluate this sum two times, one time for each possibility:
$$\begin{align}
S_{12} = 1 &\Longrightarrow S = \frac 32, \frac 12 \\
S_{12} = 0 &\Longrightarrow S = \frac 12
\end{align}$$
Note that the absolute value in the Clebsch-Gordan series forbids the possibility that $S < 0$. This makes sense since a magnitude of a spin cannot be negative.
In the end your two possibilities are $S = 3/2$ or $S = 1/2$. Due to Hund's first rule the $S = 3/2$ term is the ground state and the $S = 1/2$ terms are excited states.
Next, the $L$ in the term symbol is not equal to 1 just because the p orbitals have $l = 1$. Note that one is a small $l$ and the other is a big $L$. The big $L$ in the term symbol is obtained again by coupling the individual values of $l$ for each orbital. In this case all three p orbitals have $l =1$ and you would have to work through the Clebsch-Gordan series again.
$$L_{12} = l_1 + l_2 , l_1 + l_2 - 1, \cdots , |l_1 - l_2|$$
$$L = L_{12} + l_3, L_{12} + l_3 - 1, \cdots, |L_{12} - l_3|$$
and if you set $l_1 = l_2 = l_3$, should reach the conclusion that $L$ can take on the values of $3,2,1,0$. However you have to be careful about this, because the $L = 3$ value is forbidden by the Pauli principle, which is not taken into account by the Clebsch-Gordan series. A term with $L = 3$ would necessitate the existence of a microstate with $m_L = +3$, which would imply that all three electrons are in the p orbital with $m_l = +1$, which is not possible.
1 Technically, the spin is $\displaystyle \sqrt{\frac{1}{2}\left(\frac{1}{2} + 1\right)}\hbar$, but whatever. The correct phrasing is that the quantum number representing the magnitude of the spin can only be $1/2$.
