Proving $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\cdots\left(1+\dfrac{1}{n^3}\right)<3$ for all positive integers $n$

As mentioned in the comments, the trick to subtract something (or multiply with something $< 1$) and prove a stronger inequality than required is standard.

So we want to prove

$$p_n \leqslant 3- a_n$$

for a preferably simple $a_n > 0$ by induction. For the induction start, we here need $a_1 \leqslant 1$. For the induction step to work, we need

$$\left( 1 + \frac{1}{(n+1)^3}\right)(3-a_n) \leqslant 3 - a_{n+1}.$$

Multiplying out and cancelling the $3$, we get

$$-a_n + \frac{3}{(n+1)^3} - \frac{a_n}{(n+1)^3} \leqslant -a_{n+1}.$$

Throwing away the $\frac{a_n}{(n+1)^3}$ term to simplify the calculations, we see that

$$\frac{3}{(n+1)^3} \leqslant a_n - a_{n+1}$$

is sufficient. With the ansatz $a_n = \frac{c}{n^k}$, we have

$$a_n - a_{n+1} = c\frac{(n+1)^k-n^k}{n^k(n+1)^k} \approx c\frac{k}{n(n+1)^k},$$

so here we need $k \leqslant 2$. But $a_1 \leqslant 1$ requires $c\leqslant 1$, and hence $k = 2$ doesn't work. So we try $k = 1$ and find $a_n = \frac{1}{n}$ works.

Since: $$ 1+\frac{1}{k^3}=\left(1+\frac{1}{k}\right)\left(1-\frac{1}{k}+\frac{1}{k^2}\right) = \left(1-\frac{1}{k^2}\right)\left(1+\frac{1}{k(k-1)}\right) $$ and: $$ \prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)=\frac{1}{2} $$ we have: $$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right)=\prod_{k=2}^{+\infty}\left(1+\frac{1}{k(k-1)}\right)=\prod_{k=1}^{+\infty}\left(1+\frac{1}{k(k+1)}\right),$$ but since $1+x < e^x$ and $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ it follows that: $$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right)<\exp\sum_{k=1}^{+\infty}\frac{1}{k(k+1)}=e<3.$$ With the same trick we can also prove the stronger: $$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right) < \frac{3}{2}\sqrt{e} <\frac{5}{2}.$$