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 $