If all the 27 numbers are equal to 1, we have nothing to prove. Assume the cube contains some numbers greater than 1 and suppose there is an even number between them. If $n$ is the smallest even number written inside a cube, then $n$ should be the sum of three odd numbers, which is impossible due to parity reasons. So all the 27 numbers are odd.
We shall prove that one of the numbers in every strip is equal to 1. Assuming the contrary, there is a strip that doesn't contain 1. If $a$ is the smallest number in this strip, then $a$ should be the sum of three numbers less than $a$ and greater than 1, a contradiction with the minimality of $a$. It follows that each strip contains at least an 1, so at least 9 numbers inside the $3 \times 3 \times 3$ cube are equal to 1.

Let $a_1 \le a_2 \le \dots \le a_{18}$ the other numbers inside the big cube. If $a_1 > 3$, then one of three numbers whose sum is $a_1$ must be greater than 1 and less than $a_1$, which contradicts the choice made for $a_1$.
Therefore, $a_1 \in \{1,3\}$, so $a_2 \in \{1,3,5\}$. If $a_2 \le 3$, then $a_3 \le 1+a_1+a_2 \le 7$. If $a_2 = 5$, then $a_1$ and $a_2$ are on the same strip, so at most one of them may be used to write $a_3$ as a sum of three numbers from the cube. Hence, $a_3 \le 1+1+a_2 = 7$, so consequently $a_3 \in \{1,3,5,7\}$.

If $a_3 = 7$, it must exist a strip containing $a_2$ and $a_3$, so $a_2$ and $a_3$ cannot be simultaneously terms of expressing $a_4$ as a sum of three numbers inside the cube, so $a_4 \le a_3+a_1+1 \le 7+3+1 = 11$. If $a_3 \le 5$, then $a_4 \le a_3+a_2+a_1 \le 5+5+1 = 11$. Therefore, $a_4 \le 11$.
More general, we have $a_{k+3} \le a_{k+2} + a_{k+1} + a_k$.
If $a_{k+1}$ and $a_{k+2}$ are on a same strip, at most one of them may be used to write $a_{k+3}$ as a sum of three numbers written in the unit-cubes. Since $a_{k+2} \le a_{k+1} + a_k + a_{k-1}$, it follows that $a_{k+3} \le a_{k+2} + a_k + a_{k-1} \le a_{k+1} + 2a_k + 2a_{k-1}$.
If no strip contains $a_{k+1}$ and $a_{k+2}$, since $a_{k+2} \le a_k + a_{k-1} + a_{k-2}$, we obtain $a_{k+3} \le a_{k+2} + a_{k+1} + a_k \le a_{k+1} + 2a_k + a_{k-1} + a_{k-2} \le a_{k+1} + 2a_k + 2a_{k-1}$.
Subsequently, $a_{k+3} \le a_{k+1} + 2a_k + 2a_{k-1}$, for each $k \ge 2$. Successively, we infer that $a_5 \le 23$, $a_6 \le 35$ and $a_7 \le 59$, so at least 16 numbers written inside the unit-cubes are either smaller than 60.