There is a simpler way (in terms of manual computations) than the other answer.
We use the following facts:
1)
If $A$ is an integer then $$ A -[x] = [A-x + \delta] $$ where $\delta = 0$ if $x$ is an integer, and $\delta = 1$ otherwise.
2)
If $n > 1$ is a positive integer then $$ [\frac{[x]}{n}] = [\frac{x}{n}]$$
Using this we see that
$$[\frac{n-12 - [\frac{n-17}{25}]}{3}] = [\frac{n-12-\frac{(n-17)}{25}+\delta}{3}]$$
where $\delta = 0$ iff $n=17 \mod 25$.
$$ = [\frac{24n-283+25\delta}{75}]$$
Let $24n+39 = r \mod 75$ where $0 \leq r < 75$.
Notice that $r$ must be divisible by $3$. Thus $0 \leq r \leq 72$.
Suppose $24n+39 = 75m+r$, then we have that
$$[\frac{8n+13}{25}] = [\frac{24n+39}{75}] = m$$
and $$[\frac{24n-283+25\delta}{75}] = [\frac{75m+r-375+53+25\delta}{75}]$$
= $$m-5 + [\frac{r + 53 + 25\delta}{75}]$$
Now $\delta = 0$ iff $n=17 \mod 25$ iff $r = 72$.
Thus for $r < 72$, $\delta = 1$, for which we have
$$[\frac{r + 53 + 25\delta}{75}] = [\frac{r + 53 + 25}{75}] = 1$$
For $r = 72$, we have $\delta=0$ for which we have
$$[\frac{r + 53 + 25\delta}{75}] = [\frac{r + 53}{75}] = 1$$
Thus the whole expression is
$$ m - ((m-5)+1) = 4 $$