Let me answer the question for $p>3$ and $SL_2$. I imagine that a similar method will work for the other cases but I haven't checked.

Some notation: $G=SL_2(\mathbb{Z}/p^n\mathbb{Z})$, $Z=\{I, -I\}$ and, for $i=1,\dots, n$,
$$K_i := \ker \pi_{n,i}.$$

**Proposition**: The proper normal subgroups of $G$ are $K_i$ and $K_i\times Z$ for $i=1,\dots, n$.

**Sketch of basic steps of proof**:

If $n=1$, $G$ is quasisimple and the result is immediate. Assume from here on that $n>1$.

Observe that $K_1$ is a normal
$p$-subgroup of $G$, that
$|K_1|=p^{3(n-1)}$ and $G/K_1\cong
SL_2(\mathbb{Z}/p\mathbb{Z})$, a
quasisimple group since $p>3$.

- Observe, next that, for $i=1,\dots,
n$, |$K_i| = p^{3(n-i)}$ and,
moreover, $$\{1\}=K_n \lhd K_{n-1}
\lhd \cdots \lhd K_1$$ is a chain of
normal subgroups. In fact this is an
upper central series for the group
$K_1$, i.e $K_{i-1}/K_i = Z(K_1/K_i)$
and $K_{i-1}/K_i$ is elementary
abelian of order $p^3$.
- Now $G$ acts naturally on the group
$K_1$ by conjugation. The upper
central series structure just
described implies that this action
induces an action of $G/K_1 =
SL_2(\mathbb{Z}/p\mathbb{Z})$ on the
groups $K_{i-1}/K_i$. Thus the group
$K_{i-1}/K_i$ becomes a 3-dimensional
module for the group
$SL_2(\mathbb{Z}/p\mathbb{Z})$. It is
easy to see that for $i=2,\dots, n$,
these modules are isomorphic. Fact to
check: This module is irreducible.
- Now let $N$ be a normal subgroup of
$G$. Suppose first that $N\cap K_1$
is trivial. Then $N$ is isomorphic to
a normal subgroup of
$SL_2(\mathbb{Z}/p\mathbb{Z})$, i.e.
$N$ is trivial, equal to $Z$, or
isomorphic to
$SL_2(\mathbb{Z}/p\mathbb{Z})$. In
the latter case we would have
$G=K_1\times N$ and it is easy to
check that this does not happen. Thus
$N$ is trivial or equal to $Z$.
- Assume next that $N\cap K_1$ is
non-trivial. In particular $N\cap K_1$ is
a non-trivial normal subgroup of
$K_1$. We use the following easy
fact:
*A non-trivial normal subgroup
of a $p$-group intersects the center
of that $p$-group non-trivially.* Thus
$N\cap K_{n-1}$ is non-trivial and is
a normal subgroup of $G$. But, since
$SL_2(\mathbb{Z}/p\mathbb{Z})$ acts
irreducibly on the module $N\cap
K_{n-1}$, $N$ must contain $K_{n-1}$.
If $N\cap K_1 = K_{n-1}$, then there
are three possibilities for $N$, namely $N=
K_{n-1}$, $N=K_{n-1} \times Z$ or $N=K_{n-1}.
SL_2(\mathbb{Z}/p\mathbb{Z})$. If $n=2$, the last possibility corresponds to $G$ and we are done. If $n>2$, then the
last possibility is impossible just
as before.
- Now the proof is completed by
observing that $G/K_{n-1} \cong
SL_2(\mathbb{Z}/p^{n-1}\mathbb{Z})$ and
appealing to induction.

**Final remark**: I've read some of Bourgain & Gamburd's work dealing with $SL_2$. They tend to (implicitly) consider the center as a trivial normal subgroup as their work deals with asymptotics on $p$ which are unaffected by $Z$. This explains the apparent inaccuracy of their assertion that the $K_i$ are all of the normal subgroups of $SL_2(\mathbb{Z}/p^n\mathbb{Z})$.