= Solution
For a separated filtered group define
$$
G_\lambda=\{g:\omega(g)\geq\lambda\},
\qquad
G_{\lambda+}=\{g:\omega(g)>\lambda\}.
$$
The <associated graded Lie algebra of a filtered group> is
$$
\operatorname{gr}G=\bigoplus_\lambda G_\lambda/G_{\lambda+}.
$$
Each quotient is abelian because $[G_\lambda,G_\lambda]\subseteq G_{2\lambda}\subseteq G_{\lambda+}$. If $x\in G_\lambda$ and $y\in G_\mu$, set
$$
[\operatorname{gr}_\lambda x,\operatorname{gr}_\mu y]
=\operatorname{gr}_{\lambda+\mu}[x,y].
$$
The standard commutator identities $[xy,z]=[x,z]^y[y,z]$ and $[x,yz]=[x,z][x,y]^z$ prove well-defined bilinearity, while the Hall-Witt identity gives the <Jacobi identity>. If $\omega([x,y])=\omega(x)+\omega(y)$, this bracket has nonzero initial form, so the graded Lie algebra is nonabelian.
For a <p-valuation>, define
$$
t\operatorname{gr}_\lambda(g)=\operatorname{gr}_{\lambda+1}(g^p).
$$
The p-power axiom and the <Hall-Petrescu formula> make this well defined and turn each homogeneous component into part of a graded $\mathbb F_p[t]$-module. The leading-term congruence $[x^p,y]\equiv[x,y]^p$ modulo terms of valuation greater than $\omega(x)+\omega(y)+1$ makes the bracket $\mathbb F_p[t]$-bilinear.
For the given upper-triangular group, write an element of degree $n$ to leading order as
$$
1+p^n\begin{pmatrix}\alpha&\beta\\0&0\end{pmatrix},
\qquad \alpha,\beta\in\mathbb F_p.
$$
Let $X_n,Y_n$ denote the classes with $(\alpha,\beta)=(1,0),(0,1)$. Matrix commutators give
$$
[\alpha X_n+\beta Y_n,\gamma X_m+\delta Y_m]
=(\alpha\delta-\gamma\beta)Y_{n+m},
$$
and pth powers give $tX_n=X_{n+1}$ and $tY_n=Y_{n+1}$. Thus
$$
\boxed{\operatorname{gr}G=\mathbb F_p[t]X_1\oplus\mathbb F_p[t]Y_1,
\qquad [X_1,Y_1]=tY_1.}
$$
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