The question is very badly worded, but from the answers, I think it is meant to be taken this way.
Treat the $8\times 8\times 8$ cube as a cake.
Given one of the smaller cubes, we cut the cake along three planes in three different directions, so that the planes are extensions of the faces of the smaller cube.
The smaller cube is labeled with the smallest piece of cake obtained by such cuts which contains the smaller cube.
If you label the smaller cubes as triples $(i,j,k)$ with $1\leq i,j,k\leq 8$, then the cube would be labeled:
$$v(i,j,k)=\min(i,9-i)\min(j,9-j)\min(k,9-k)$$
That's because your cut piece will have either $i$ or $9-i$ for the edge length, depending on which side of it you cut, and same for $j,k$.
Now (1) amounts to wanting:
$$v(1,1,1)+v(2,2,2)+\cdots+v(8,8,8)=1^2+2^3+3^3+4^3+4^3+3^3+2^2+1^3$$
as the answer above states.
For (2), you want at least one of $\min(i,9-i),\min(j,9-j),\min(k,9-k)$. This means tat either $i=3,6$ or $j=3,6$, or $k=3,6$.
It's easier to count the ones that are not of this form - that is, where $i,j,k\in\{1,2,4,5,7,8\}$ then there are $6^3=216$ such triples, and thus $8^3-6^3=296$.
For (3), again, the phrasing is odd. What the answer seems to compute is the sum of the values on the faces of the cubes on the surface, not the sum of the values on the cubes on the surface. So, the corners for example have three faces on the surface, and thus are counted three times.
The formula for $v$ lets us see that we need to factor $8$ into three values no bigger than $4$, which means $8=4\cdot 2\cdot 1$ or $8=2\cdot2\cdot 2.$ You'll have to count these cases up sepearately.
Let's define some coordinates first. Say the cube is $[0, n]^3$ (so vertices at $\{(x, y, z) \in \mathbb R^3 : x, y, z \in \{0, n\}\}$), where $n \in \mathbb N$ is the number of subdivisions you want (in your case $25$, since $25^3 = 15625$). The smaller cubes have their vertices on integer lattice points whose coordinates are between $0$ and $n$. Specifically, for each lattice point $(a, b, c)$ with $a, b, c \in \{0, \dotsc, n-1\}$ there is a small cube
$$ [a, a+1] \times [b, b+1] \times [c, c+1] \,. $$
We'll call $(a, b, c)$ the "low corner" of this cube, and $(a+1, b+1, c+1)$ the "high corner".
In order to find a cross section of the large cube which is a regular hexagon, we take the intersection with the plane $x + y + z = 1.5n$. This plane divides the cube in half, intersecting the cube's "frame" at the midpoints of six edges. I believe this is the type of cross section you had in mind.
In order to count how many small cubes are cut, we just need to count how many small cubes have their "low corner" $(a, b, c)$ and "high corner" $(a+1, b+1, c+1)$ on opposite sides of this plane. This occurs when
$$ a + b + c \leq 1.5n \leq (a+1) + (b+1) + (c+1) \,,$$
or equivalently
$$ a + b + c \in [1.5n-3, 1.5n] \,. $$
Since $a+b+c$ is an integer, all we need to do is count the number of lattice points $(a, b, c)$ with $a, b, c \in \{0, \dotsc, n-1\}$ such that
$$ a + b + c \in \{\lceil1.5n\rceil-3, \dotsc, \lfloor1.5n\rfloor\} \,. $$
This means the number of small cubes cut is
$$ C(n) = \sum_{s=\lceil1.5n\rceil-3}^{\lfloor1.5n\rfloor} S_3(s; n) \,, \tag{$*$} $$
where $S_3(s; n)$ is defined as the number of ways to sum $3$ nonnegative integers less than $n$ to get $s$. (Define $S_k(s; n)$ similarly for other $k$.)
This $S_k(s; n)$ might have some well-known name, but I am not aware of it. Either way, we can derive it by applying the approach from this answer. First of all, the total number of ways of summing $k$ nonnegative integers to get $s$ is
$$ S_k(s) = \begin{cases} \binom{s+k-1}{k-1} &:\, s \geq 0 \\ 0 &:\, \text{otherwise} \end{cases} \,. $$
Among these, the number of ways where the first number is $\geq n$ is
$$ S_k(s-n) \,, $$
and likewise for the number of ways where the second number is $\geq n$, etc. The number of ways where the first two (or indeed, any particular two) numbers are $\geq n$ is
$$ S_k(s-2n) \,. $$
In general, if we require $j$ of the numbers to be $\geq n$, the number of ways of summing to get $s$ is
$$ S_k(s-jn) \,. $$
Using Inclusion-Exclusion, the total number of ways of summing to get $s$ where at least one of the summands is $\geq n$, is
$$ \sum_{j=1}^k (-1)^{j-1} \binom{k}{j} S_k(s-jn) \,, $$
and thus
\begin{align*}
S_k(s; n)
&= S_k(s) - \sum_{j=1}^k (-1)^{j-1} \binom{k}{j} S_k(s-jn) \\
&= \sum_{j=0}^k (-1)^j \binom{k}{j} S_k(s-jn) \,.
\end{align*}
Now, let's apply this result back to $(\ast)$. We wanted to compute $S_3(s; n)$, but only cared about cases where $s \leq \lfloor1.5n\rfloor < 2n$. Assuming $s < 2n$, the formula for $S_3(s; n)$ simplifies to
$$ S_3(s) - 3 S_3(s-n) \,, $$
so the final answer for the number of small cubes cut is
\begin{align*}
C(n)
&= \sum_{s=\lceil1.5n\rceil-3}^{\lfloor1.5n\rfloor} \big(S_3(s) - 3 S_3(s-n)\big) \\
&= \sum_{s=\lceil1.5n\rceil-3}^{\lfloor1.5n\rfloor} S_3(s) - 3 \sum_{s=\lceil0.5n\rceil-3}^{\lfloor0.5n\rfloor} S_3(s) \\
&= \sum_{s=\max(0, \lceil1.5n\rceil-3)}^{\lfloor1.5n\rfloor} \binom{s+2}{2} - 3 \sum_{s=\max(0, \lceil0.5n\rceil-3)}^{\lfloor0.5n\rfloor} \binom{s+2}{2} \\
&= \frac12 \sum_{s=\max(2, \lceil1.5n\rceil-1)}^{\lfloor1.5n\rfloor+2} s(s-1) - \frac32 \sum_{s=\max(2, \lceil0.5n\rceil-1)}^{\lfloor0.5n\rfloor+2} s(s-1) \,.
\end{align*}
Edit (from comments): if $n$ is odd and $\geq 5$, we can instead simplify the expression to
\begin{align*}
C(n)
&= \sum_{s=(3n-5)/2}^{(3n-1)/2} S_3(s) - 3 \sum_{s=(n-5)/2}^{(n-1)/2} S_3(s) \\
&= \sum_{s=(3n-1)/2}^{(3n+3)/2} \frac{s(s-1)}{2} - 3 \sum_{s=(n-1)/2}^{(n+3)/2} \frac{s(s-1)}{2} \\
&= \frac{(3n-1)(3n-3) + (3n+1)(3n-1) + (3n+3)(3n+1)}{8} \\
&\quad - 3\cdot \frac{(n-1)(n-3) + (n+1)(n-1) + (n+3)(n+1)}{8} \\
&= \frac{27n^2+5}{8} - \frac{9n^2+15}{8} \\
&= \frac{9n^2-5}{4} \,.
\end{align*}
In fact, this expression also holds for $n=1$ and $n=3$ (see below).
In the original problem, $n$ was $25$. Plugging this in, we get an answer of
$$ C(25) = (9\cdot25^2 - 5)/4 = \boxed{1405} \,. $$
As a sanity check, let's also compute $C(1)$, $C(2)$, and $C(3)$:
\begin{align*}
C(1) &= \frac12 \sum_{s=2}^3 s(s-1) - \frac32 \sum_{s=2}^2 s(s-1) = 1 \,, \\
C(2) &= \frac12 \sum_{s=2}^5 s(s-1) - \frac32 \sum_{s=2}^3 s(s-1) = 8 \,, \\
C(3) &= \frac12 \sum_{s=4}^6 s(s-1) - \frac32 \sum_{s=2}^3 s(s-1) = 19 \,.
\end{align*}
These numbers match what I calculated by hand.
Best Answer
If you remove the outer layer, it means every single side became shorter by exactly $2$.
So you have a cube with a side of $8$ in hand, and its volume is indeed $$8^3 = 512.$$