You calculations should become simpler if you consider the semidirect product $G := \left(\begin{array}{cccc} 1 & * & * & *\\ & 1 & * & *\\ & & 1 & *\\ & & & 1 \end{array}\right) = \left(\begin{array}{cccc} 1 & & * & *\\ & 1 & * & *\\ & & 1 & \\ & & & 1 \end{array}\right)\rtimes \left(\begin{array}{cccc} 1 & * & & \\ & 1 & & \\ & & 1 & *\\ & & & 1 \end{array}\right) =: N\rtimes H.$
Both $N$ and $H$ are elementary abelian, and you have mostly to understand the conjugation action of the three nonzero elements of $H$ on $N$.
EDIT: As you finished your assignment I now post an almost complete solution.
Denoting the elements of $H$ as $1, x, y$ and $z$ we get: $x^{-1}\cdot\left(\begin{array}{cc} b & a \\ d & c \end{array}\right)\cdot x := \left(\begin{array}{cccc} 1 & 1 & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{array}\right)\cdot\left(\begin{array}{cccc} 1 & & b & a\\ & 1 & d & c\\ & & 1 & \\ & & & 1 \end{array}\right)\cdot\left(\begin{array}{cccc} 1 & 1 & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{array}\right) = \left(\begin{array}{cccc} 1 & & b+d & a+c\\ & 1 & d & c\\ & & 1 & \\ & & & 1 \end{array}\right) = \left(\begin{array}{cc} b+d & a+c\\ d & c \end{array}\right)$
and (only in short notation) $y^{-1}\cdot\left(\begin{array}{cc} b & a \\ d & c \end{array}\right)\cdot y = \left(\begin{array}{cc} b & a+b\\ d & c+d \end{array}\right)$
$N$ being normal and $H$ being abelian, the conjugacy classes are contained in the cosets $hN$ for $h \in H$. As $N$ is abelian, one can determine the centralizers of the elements of $N$ by looking at the action of $x$ and $y$ on $N$. One obtains that $N$ contributes $2\cdot 1 + 3\cdot 2 + 2\cdot 4$ to the class equation of $G$.
Now consider the centralizers/conjugacy classes of $X := \langle x, N\rangle$: $N$ is abelian and has index $2$ in $X$. By the action of $x$ on $N$ is $C := \left(\begin{array}{cc} * & *\\ 0 & 0 \end{array}\right) = \mathrm{Z}(X)$ the center of $X$, and $xC$ is the conjugacy class of $x$ in $X$. As $N$ is abelian, the elements $t$ of $X\setminus N$ all have centralizers in $X$ of order $4$ and their conjugacy classes (in $X$) are the cosets $tC$.
As $X$ has index $2$ in $G$ (i.e., is normal of prime index), for elements $t \in X\setminus N$ either the centralizer grows by factor $2$ or its conjugacy class is merged with another conjugacy class.
As the centralizer of $y$ in $G$ is $\langle x, \left(\begin{array}{cc} 0 & *\\ 0 & * \end{array}\right)\rangle$, there are two conjugacy classes (of $G$) of order $4$ contained in $xN$ ("growing centralizer"). As $y^{-1}\cdot x\left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)\cdot y = x\left(\begin{array}{cc} 0 & 0\\ 1 & 1 \end{array}\right)$ the other two conjugacy classes merge to a conjugacy class of $G$ of order $8$ contained in $xN$. Hence, $xN$ contributes $2\cdot 4 + 8$ to the class equation of $G$.
By symmetry, i.e., using automorphisms of $G$ fixing $N$ that map $x$ to $y$ (easy) rsp. $x$ to $z$ (not too difficult), $yN$ and $zN$ contribute the same as $xN$ does to the class equation, which is $ G = 2\cdot 1 + 3\cdot 2 + 2\cdot 4 + 3\cdot[2\cdot 4 + 8],$ as Ben already commented.
Extra credit: Find the class equation for $\mathbb{F}_2$ generalized to any finite field $\mathbb{F}_q$.
(Hint: Conjugate with diagonal matrices)
EDIT 2: The action of the element $x$ on $N$ has a two-dimensional centralizer $C_N(x)$ (viewing the elementary abelian group as $\mathbb{F}_2$-vector space) that equals the commutator $[x, N]$. The same holds for $y$ and $z$, and the centralizers of any pair intersect one-dimensionally. As $N$ is elementary abelian, one can find an automorphism of $G$ exchanging $x$ and $z$ and fixing $N$ (not pointwise, but as a set):
One automorphism exchanging $x$ and $z$ maps a basis of $N$ like this:
$\left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right)\mapsto \left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right), \left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)\mapsto \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$, whereas $\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right)$ and $\left(\begin{array}{cc} 0 & 0\\ 0 & 1 \end{array}\right)$ are fixed.