How does one find the ground-state term symbol for a configuration that is exactly half-filled?
The full calculation is laid out below. Start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number $s = 1/2$ and magnetic quantum number  $m_s = \pm 1/2$. Orbitals have angular momentum of $s=0, p=1, d=2, f=3$ etc.

The total spin angular momentum is the series of values
$$S = |s_1 + s_2|\cdots|s_1 − s_2|$$
where $s_1 , s_2 , s_3 =+1/2$  are the quantum numbers of each electron. The total angular momentum is 
$$ L=|l_1+l_2|\cdots|l_1−l_2|$$
and total angular momentum 
$$J=|L+S|\cdots|L−S|$$ 
Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by $1$. 

The term symbol has the form $^{2S+1}L_J$. The super-prefix is the spin multiplicity, for spin angular momentum S this is $2S+1$ or in general $2X+1$ for angular momentum X. 

When there are three 3 electrons the the spin and orbital angular momentum terms have to be added in two parts. First as a pair, with the equations above then again with each of the values in the series with the last electron or orbital.

For the spin the S values are
$S = |s_1 + s_2|\cdots|s_1 − s_2| = |1/2+1/2|\cdots|1/2-1/2| = 1, ~ 0 $ . 
We call these values $S_1$ and $ S_0$
To calculate the total with the third electron gives 
$$S_{s_1}=|S_1+1/2|\cdots|S_1 - 1/2| = 3/2 , ~ 1/2$$
and for the $S_0$ there is one value $S_0=1/2$. Thus for the spin the values are $3/2$ and $1/2$. This means that the multiplicity produces quartet and doublet states, thus so far we have $^4L_J$ and $^2L_J$.

The same method is followed for the orbital angular momentum. The first two p orbitals give 
$$ L=|l_1+l_2|\cdots|l_1−l_2|= |1+1|\cdots|1-1| = 2,~1,~0 $$
then for each of these values called $L_1,L_2,L_3$ we combine with the last orbital angular momentum, for example
$$ L_{L_1}=|L_1+l_1|\cdots|L_1−l_2|= |2+1|\cdots|2-1| = 3,~2,~1 $$
and so forth for the other values. The L values are $3,2,1,0$

Putting all these values into a table gives 
$$\begin{matrix} L &amp; S &amp;  J &amp;  \text{term symbol}\\ 
3 &amp; 3/2 &amp; 9/2, 7/2, 5/2, 3/2 &amp; ^4F_{9/2, 7/2, 5/2, 3/2} \\
2 &amp; 3/2 &amp; 7/2, 5/2, 3/2,1/2 &amp; ^4D_{ 7/2, 5/2, 3/2,1/2}\\
1 &amp; 3/2 &amp; 5/2, 3/2,1/2 &amp; ^4P_{5/2, 3/2,1/2}\\
0 &amp; 3/2 &amp; 3/2 &amp; ^4S_{3/2}\\
3 &amp; 1/2 &amp; 7/2, 5/2 &amp; ^2F_{7/2,5/2}\\
2 &amp; 1/2 &amp; 5/2, 3/2 &amp;^2D_{5/2, 3/2}\\
1 &amp; 1/2 &amp; 3/2, 1/2 &amp; ^2P_{3/2, 1/2}\\
0 &amp; 1/2 &amp; 1/2 &amp; ^2S_{1/2}\\
\end{matrix}$$
Finally to find the lowest energy term use Hund’s rules, i.e. of the terms given by equivalent electrons, i.e. having the same L values, the one with the greatest multiplicity is lowest in energy and of these the lowest is that one with the greatest L value. The only terms remaining after considering Pauli exclusion are $^4S$ which is lowest in energy and next $^2D$ then $^2P$