Let $n$ be some fixed positive integer.
Since $[0,n+1]$ and $f_n(x)$ are symmetric about $x=\frac{n+1}2$, the minimum value of $f_n$ over $[0, n+1]$ is the same as the minimum of $f_n$ over $[0, \frac{n+1}{2}]$. From now on, we will assume $f_n$ is defined on $[0, \frac{n+1}2]$. It is immediate to verify $f_1(x)=0$. From now on, we will assume $n>1$.
As observed in the question, $(\sum_{i=1}^{n}|x-i|)^2$ is not easy to deal with because of the absolute values. The most common way to remove absolute values is, well, to separate the domain of the variable into small pieces so that we know how to take the absolute value in each piece.
Let $\lfloor x\rfloor$ be the integer part of $x$, i.e., $\lfloor x\rfloor\in\mathbb N$ and $0\le x - \lfloor x\rfloor\lt1$. There are two cases for $\lfloor x\rfloor$.
- $\lfloor x\rfloor=0$. $\quad( \sum_{i=1} ^ {n} | x-i | )^2=( \sum_{i=1} ^ {n} (i-x) )^2=( \frac{n(n+1)}2-nx)^2.$
- $1\le \lfloor x\rfloor\le \frac{n+1}2$. $\quad( \sum_{i=1} ^ {n} | x-i | )^2=( \sum_{i=1} ^ {\lfloor x\rfloor} (x-i)+\sum_{i=\lfloor x\rfloor+1} ^ {n} (i-x) )^2\\=\left( \sum_{i=1} ^ {\lfloor x\rfloor} x+\sum_{i=\lfloor x\rfloor+1} ^ {n} (-x)+\sum_{i=1} ^ {\lfloor x\rfloor} (-i)+\sum_{i=\lfloor x\rfloor+1} ^ {n}i \right)^2\\=\left( (2\lfloor x\rfloor-n)x-\frac{\lfloor x\rfloor(\lfloor x\rfloor+1)}2+\frac{(n-\lfloor x\rfloor)(n+\lfloor x\rfloor+1)}2\right)^2.$
Note the formula above for the second case holds for the first case, too.
As computed in the question, we have.
$$\sum_{i=1}^{n}(x-i)^2=nx^2-n(n+1)x+\dfrac{n(n+1)(2n+1)}{6}.$$
So,
$$\begin{aligned}
&\quad\quad f_n(x)=\sum_{i=1} ^ {n}| x-i | )^2 - \sum_{i=1} ^{n} (x-i)^2\\
&=\left( (2\lfloor x\rfloor-n)x-\frac{\lfloor x\rfloor(\lfloor x\rfloor+1)}2+\frac{(n-\lfloor x\rfloor)(n+\lfloor x\rfloor+1)}2\right)^2 - \left(nx^2-n(n+1)x+\dfrac{n(n+1)(2n+1)}{6}\right)\\
&=-nx^2+(n^2+2\lfloor x\rfloor)x+ \text{ (some formula that depends only on } \lfloor x\rfloor\text{ and }n)\\
&=-n\left(x-(\frac n2+\frac{\lfloor x\rfloor}n)\right)^2+ \text{ (some formula that depends only on } \lfloor x\rfloor\text{ and }n).\\
\end{aligned}$$
Since $\left\lfloor\lfloor x\rfloor\right\rfloor=\lfloor x\rfloor$, we have
$$f_n(x)-f_n(\lfloor x\rfloor,n)=-n\left(x-(\frac n2+\frac{\lfloor x\rfloor}n)\right)^2+n\left(\lfloor x\rfloor-(\frac n2+\frac{\lfloor x\rfloor}n)\right)^2=n(x-\lfloor x\rfloor)(n+\frac{2\lfloor x\rfloor}n-x-\lfloor x\rfloor).
$$
- If $x\le\frac n2$, we have $n+\frac{2\lfloor x\rfloor}n-x-\lfloor x\rfloor\ge n -x-x\ge0.$
- Otherwise $x\gt\frac n2.$ Recall that $x\le\frac{n+1}2$.
- $n$ is even. Then $\lfloor x\rfloor=\frac n2$. We have $n+\frac{2\lfloor x\rfloor}n-x-\lfloor x\rfloor= \frac{n+1}2- x + \frac12\gt0.$
- $n$ is odd.
- If $x=\frac {n+1}2$, i.e., $x$ is an integer, $x-\lfloor x\rfloor=0$.
- Otherwise, suppose $x<\frac {n+1}2$. Then $\lfloor x\rfloor=\frac{n-1}2$, and $n+\frac{2\lfloor x\rfloor}n-x-\lfloor x\rfloor=1-\frac 1n+\frac{n+1}2-x \gt0.$
Combining all cases, we have $f_n(x)-f_n(\lfloor x\rfloor,n)\ge0$. That is, if $f_n(\cdot)$ reaches its minimum at $x$, then it must also reach
its minimum at $\lfloor x\rfloor$, an integer.
So, in order to find the minimum of $f_n(x)$, we can restrict $x$ to integers. An easy way to obtain $\sum_{n=1}^{11} (-1)^{n+1} a_n$ should be computing $f_n(x)$ at every possible integer $x$ by brute force, by some programming using your favourite programming language.
Let us check where $f_n(x)$ can reach its minimum. Assume $f_n(x)$ is defined on integers in $[0, \frac{n+1}2]$.
$$\begin{aligned}
&\quad\quad f_n(x)\\
&=\left( (2x-n)x-\frac{x(x+1)}2+\frac{(n-x)(n+x+1)}2\right)^2 - \left(nx^2-n(n+1)x+\dfrac{n(n+1)(2n+1)}{6}\right)\\
&=\left(x^2-(n+1)x+ \frac{n^2+n}2\right)^2-\left(nx^2-n(n+1)x+\dfrac{n(n+1)(2n+1)}{6}\right).\\
\end{aligned}$$
Suppose $x+1\le \frac{n+1}2$ so that $f_n(x+1)$ is defined.
$$\begin{aligned}
&\quad\quad f_n(x) - f_n(x+1)\\
&=\left( (-(x+x+1)+(n+1))(x^2+(x+1)^2-(n+1)(x+x+1)+n^2+n)\right)+\left(n(x+x+1)-n(n+1)\right)\\
&=(n-2x)(2x^2-2nx+n^2)+(2nx-n^2)\\
&=(n-2x)\left(2(x-\frac n2)^2+\frac {n(n-2)}2\right).\\
&\ge0
\end{aligned}$$
That means, $f_n(0), f_n(1), f_n(2), ...$ is a non-increasing sequence. The last item of the sequence is either $f_n(\frac n2)$ for even $n$ or $f_n(\frac {n+1}2)$ for odd $n$. So,
$$ a_n=
\begin{cases}
f_n(\frac {n+1}2)\quad \text{ if } n \text{ is odd,}\\
f_n(\frac n2)\quad\quad \text{ if } n \text{ is even.}
\end{cases} $$
Now let us compute $a_n$ into closed formula.
For odd n,
$$\begin{aligned}
a_n
&=\left((\frac {n+1}2)^2-(n+1)\frac {n+1}2+ \frac{n^2+n}2\right)^2-\left(n(\frac {n+1}2)^2-n(n+1)\frac {n+1}2+\dfrac{n(n+1)(2n+1)}{6}\right)\\
&=\frac{(n^2-1)^2}{16}-\frac{n^3-n}{12}.\\
\end{aligned}$$
So, we have $a_{1}=0$, $a_{3}=2$, $a_{5}=26$, $a_{7}=116$, $a_{9}=340$, $a_{11}=790$.
For even $n$,
$$\begin{aligned}
a_n
&=\left((\frac n2)^2-(n+1)\frac n2+ \frac{n^2+n}2\right)^2-\left(n(\frac n2)^2-n(n+1)\frac n2+\dfrac{n(n+1)(2n+1)}{6}\right)\\
&=\frac{n^4}{16}-\frac{n^3+2n}{12}.\\
\end{aligned}$$
So, we have $a_{2}=0$, $a_{4}=10$, $a_{6}=62$, $a_{8}=212$, $a_{10}=540$.
Finally, we obtain $$\sum_{n=1}^{11} (-1)^{n+1} a_n=0-0 + 2-10 + 26-62 + 116-212 + 340-540 + 790=450.$$
