Yes, it is always possible.
Let $k$ be the smallest positive integer such that $20k^2 > n^2$. We will start with a $2k\times 2k$ axis-aligned square, which is a little larger than we need. From there, we'll need to remove a total area of $A/5$, where $A$ is a nonnegative integer equal to at most $5(4k^2-(2k-1)^2) = 20k-5$.
Our plan for removing this area is to make a single cut of slope $s/5$ for $0\le s < 15$ connecting the north and west sides of the square (as in Penguino's diagram for $n=11$), and translate our starting square along the north edge of the square until we have a $1\times s/5$ triangle. The remaining three cuts, with slope $\pm 1$, can be made at whatever integer depth we want, and will safely align with the notches no matter where they go.
Suppose the depth of the other three cuts are $a, b,$ and $c$ respectively, each at most $k$ (so that the cuts don't intersect each other). (We also need $k\ge 3$ to make sure they don't intersect the $1\times s/5$ triangle.) Then we want to obtain
$$\frac A5 = \frac{s}{10}+\frac{a^2+b^2+c^2}{2}$$
$$2A = s+5(a^2+b^2+c^2)$$
by choosing $s\in\{0,1,2,3,\ldots,14\}$, $a,b,c\in\{0,1,2,\ldots,k\}$.
First, let's see that all positive integers are expressible in the form $s+5(a^2+b^2+c^2)$ if $a,b,c$ are unrestricted. Given a target $t$, we subtract a value of $s$ with the right congruence class and express the integer $\frac{t-s}{5}$ as the sum of three squares. The only way this can fail is if either $t-s$ is negative, in which case $t<s$ and we could have simply chosen $s=t, a=b=c=0$, or if $\frac{t-s}{5}$ is not expressible as a sum of three squares, ie is of the form $4^a(8b+7)$ by Legendre's three-square theorem. But since no three consecutive integers are of this form, one of our three possible candidates for $0\le s<15$ will succeed.
So, if $2A\le 5k^2$, we'll have no way to solve this equation by choosing any of $a,b,c$ to be larger than $k$, which means that the solution we know exists must do so in our desired ranges for $a,b,c$. Since we know $A\le 20k-5$, it will suffice for
$$20k-5\le 5k^2\iff k^2-4k+1\ge 0 \iff (k-2)^2 \ge 3 \iff k-2 \ge 2 \iff k\ge 4$$
so we need $n$ large enough that $\left\lceil \frac n{\sqrt{20}}\right\rceil \ge 4\iff \frac{n}{\sqrt{20}}>3 \iff n>3\sqrt{20}\approx 13.4$. Since other users have already solved the cases for $n\le 13$, this completes the proof.
