Construction: Choose three pairwise parallel lines $l_A, l_B, l_C$ forming an infinite equilateral triangle prism (with side larger than 1). Split the $n$ fireflies among the lines as equally as possible, and say that two fireflies are friends iff they lie on different lines.
To see this works:
1. Reflect $l_A$ and all fireflies on $l_A$ in the plane containing $l_B$ and $l_C$.
2. Reflect $l_B$ and all fireflies on $l_B$ in the plane containing $l_C$ and $l_A$.
3. Reflect $l_C$ and all fireflies on $l_C$ in the plane containing $l_A$ and $l_B$.
...

Proof: Consider a valid configuration of fireflies. If there is no 4-clique of friends, then by Turán's theorem, there are at most $f(n)$ pairs of friends.
Let $g(n)$ be the answer, given that there exist four pairwise friends (say $a, b, c, d$). Note that for a firefly to move, all its friends must be coplanar.

**Claim (No coplanar $K_4$) —** We can't have four coplanar fireflies which are pairwise friends.
Proof. If we did, none of them could move (unless three are collinear, in which case they can't move). □

Claim (Key claim — tetrahedrons don't share faces often) — There are at most 12 fireflies $e$ which are friends with at least three of $a, b, c, d$.
Proof. First denote by $A, B, C, D$ the locations of fireflies $a, b, c, d$. These four positions change over time as fireflies move, but the tetrahedron $ABCD$ always has a fixed shape, and we will take this tetrahedron as our reference frame for the remainder of the proof.
WLOG, will assume that $e$ is friends with $a, b, c$. Then $e$ will always be located at one of two points $E_1$ and $E_2$ relative to $ABC$, such that $E_1ABC$ and $E_2ABC$ are two congruent tetrahedrons with fixed shape. We note that points $D, E_1$, and $E_2$ are all different: clearly $D \neq E_1$ and $E_1 \neq E_2$. (If $D = E_2$, then some fireflies won't be able to move.)
Consider the moment where firefly $a$ moves. Its friends must be coplanar at that time, so one of $E_1, E_2$ lies in plane $BCD$. Similar reasoning holds for planes $ACD$ and $ABD$.
So, WLOG $E_1$ lies on both planes $BCD$ and $ACD$. Then $E_1$ lies on line $CD$, and $E_2$ lies in plane $ABD$. This uniquely determines $(E_1, E_2)$ relative to $ABCD$:

* $E_1$ is the intersection of line $CD$ with the reflection of plane $ABD$ in plane $ABC$.
* $E_2$ is the intersection of plane $ABD$ with the reflection of line $CD$ in plane $ABC$.
Accounting for WLOGs, there are at most 12 possibilities for the set $\{E_1, E_2\}$, and thus at most 12 possibilities for $E$. (It's not possible for both elements of one pair $\{E_1, E_2\}$ to be occupied, because then they couldn't move.) □
Thus, the number of friendships involving exactly one of $a, b, c, d$ is at most $(n-16) \cdot 2+12 \cdot 3 = 2n+4$, so removing these four fireflies gives
$$g(n) \le 6 + (2n + 4) + \max\{f(n-4), g(n-4)\}.$$
The rest of the solution is bounding. When $n \ge 24$, we have $(2n+10)+f(n-4) \le f(n)$, so
$$g(n) \le \max\{f(n), (2n+10) + g(n-4)\} \quad \forall n \ge 24.$$
By iterating the above inequality, we get
$$g(n) \le \max \left\{ f(n), (2n+10) + (2(n-4)+10) \right. \\ \left. \qquad + \cdots + (2(n-4r)+10) + g(n-4r-4) \right\},$$
where $r$ satisfies $n - 4r - 4 < 24 \le n - 4r$.
Now
$$\begin{aligned} & (2n + 10) + (2(n - 4) + 10) + \dots + (2(n - 4r) + 10) + g(n - 4r - 4) \\ &= (r + 1)(2n - 4r + 10) + g(n - 4r - 4) \\ &\le \left(\frac{n}{4} - 5\right) (n + 37) + \binom{24}{2}. \end{aligned}$$
This is less than $f(n)$ for $n \ge 70$, which concludes the solution.