Here is a proof that the inequality holds for $n\geq 9$. I let you verify the remaining cases $n\leq 8$.

Let $x:=p_n$ so that $x\geq 23$ by $n\geq 9$. Then the inequality can be rewritten as
$$ 2\bigl(\pi(x)-2\bigr)<\left\lfloor\frac{x}{6}\right\rfloor x \prod_{3\leq p\leq x}\frac{p-1}{p}.$$
By increasing the LHS and decreasing the RHS, we get the following stronger inequality:
$$ 2\pi(x)<\frac{(x-5)x}{6}\prod_{3\leq p\leq x}\frac{p-1}{p}.$$
We shall further increase the LHS and decrease the RHS by using the bounds in the famous paper of Rosser-Schoenfeld (1961): Approximate formulas for some functions of prime numbers. Using (3.6) and (3.27) in this paper, we see that it suffices to show
$$ \frac{8 x}{3\log x}<\frac{(x-5)x}{6\log x}\left(1-\frac{1}{\log^2 x}\right),$$
that is,
$$ 16 < (x-5) \left(1-\frac{1}{\log^2 x}\right).$$
However, this last inequality is clear by $x\geq 23$. Done.

**Added.** The OP mentioned in a comment the following related problem. This can be solved in much the same way, using that
$$ \left(\frac{p-1}{p}\right)^3<\frac{p-2}{p}. $$
So, unless $x$ is very small, the sharper inequality in the above related problem is also true.