How to factorize $5$ in $\mathbb{Z}[\root 3 \of 2]$?

We have

$$ \mathbf Z[\sqrt[3]{2}]/(5) \cong \mathbf Z[x]/(x^3 - 2, 5) \cong \mathbf Z_5[x]/(x^3 - 2) \cong \mathbf Z_5[x]/(x+2) \times \mathbf Z_5[x]/(x^2 + 3x + 4) $$

so that the ideal $ (5) $ factors as $ (5) = \mathfrak p_1 \mathfrak p_2 $. To find the ideals $ \mathfrak p_1 $ and $ \mathfrak p_2 $, note that they correspond to the maximal ideals in the ring $ \mathbf Z_5[x]/(x^3 - 2) $. We therefore have the following factorization:

$$ (5) = (\sqrt[3]{2} + 2, 5)(\sqrt[3]{4} + 3 \sqrt[3]{2} + 4, 5) $$

Since the ring $ \mathbf Z[\sqrt[3]{2}] $ is a principal ideal domain, these ideals are principally generated by prime elements of norm $ 5 $ and $ 25 $, respectively. Some easy trial and error by hand with the norm form yields that the first ideal is generated by $ 2\sqrt[3]{4} - 3 $, and dividing $ 5 $ by this number finally gives the factorization

$$ 5 = (2\sqrt[3]{4} - 3)(6 \sqrt[3]{4} + 8 \sqrt[3]{2} + 9) $$

Here's some more detail on the trial and error part: we know that $ \mathfrak p_1 $ is a prime ideal of norm $ 5 $, and we know that $ (5) = \mathfrak p_1 \mathfrak p_2 $. Since any number with norm $ 5 $ generates a prime ideal, we can conclude by unique factorization of ideals that any such element would have to generate $ \mathfrak p_1 $. After that, we just look for solutions to the equation $ x^3 + 2y^3 + 4z^3 - 6xyz = 5 $ (the norm form), and it is easily seen that $ x = -3 $ and $ z = 2 $ do the trick.

Note that the factorization you are asking for is impossible: the prime $ 5 $ does not factor as the product of three prime ideals, but as two.

This answer is a try to use the number $1+\sqrt[3]{4}$ as OP did.

As proved by @Starfall, the prime $5$ factors as a product of two primes. With a help from SAGE, the prime factors as $$ 5 = (1+\sqrt[3]{4})(1+2\sqrt[3]{2} - \sqrt[3]{4}). $$ One might wonder why the factors look different from @Starfall's answer. In fact, we have $$ 2\sqrt[3]{4} -3 = (1+\sqrt[3]{4})/(1+\sqrt[3]{2}+\sqrt[3]{4})^2, $$ with $1+\sqrt[3]{2}+\sqrt[3]{4}$ being a fundamental unit (i.e. a generator of the free part of unit group) in $\mathbb{Z}[\sqrt[3]{2}]$. As $\mathbb{Z}[\sqrt[3]{2}]$ is a PID, it is also a UFD. This means each prime ideal is generated by a prime element in the ring. So, the factors $1+\sqrt[3]{4}$ and $1+2\sqrt[3]{2} - \sqrt[3]{4}$ are prime elements in $\mathbb{Z}[\sqrt[3]{2}]$.