1. We need to evaluate $S_n=\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2.$

The key is to understand the sign pattern of $(-1)^{\frac{k(k+1)}{2}}.$

2. Let $T_k=\frac{k(k+1)}{2}.$ We only care whether $T_k$ is even or odd.

Checking small values:

• $k=1$: $T_1=1$ odd $\Rightarrow$ sign $=-1$
• $k=2$: $T_2=3$ odd $\Rightarrow$ sign $=-1$
• $k=3$: $T_3=6$ even $\Rightarrow$ sign $=+1$
• $k=4$: $T_4=10$ even $\Rightarrow$ sign $=+1$

So the signs repeat in blocks of $4$ as $-,-,+,+,-,-,+,+,\dots$ Hence for each block of four terms, $-(4m+1)^2-(4m+2)^2+(4m+3)^2+(4m+4)^2.$

3. Since the upper limit is $4n$, group terms in blocks: $S_n=\sum_{m=0}^{n-1}\left[-(4m+1)^2-(4m+2)^2+(4m+3)^2+(4m+4)^2\right].$

Now simplify one block: \begin{align*} B_m&=-(4m+1)^2-(4m+2)^2+(4m+3)^2+(4m+4)^2. \end{align*}

Expand: \begin{align*} (4m+1)^2&=16m^2+8m+1,\ (4m+2)^2&=16m^2+16m+4,\ (4m+3)^2&=16m^2+24m+9,\ (4m+4)^2&=16m^2+32m+16. \end{align*}

Therefore, \begin{align*} B_m&=-(16m^2+8m+1)-(16m^2+16m+4)\ &\quad +(16m^2+24m+9)+(16m^2+32m+16)\ &=32m+20. \end{align*}

4. Hence $S_n=\sum_{m=0}^{n-1}(32m+20).$ So, \begin{align*} S_n&=32\sum_{m=0}^{n-1}m+20n\ &=32\cdot \frac{(n-1)n}{2}+20n\ &=16n(n-1)+20n\ &=16n^2+4n\ &=4n(4n+1). \end{align*}

Thus $S_n=4n(4n+1).$

5. Now test each option.

Option A: 1056

We solve $4n(4n+1)=1056.$ Let $x=4n$. Then $x(x+1)=1056.$ So $32\cdot 33=1056,$ thus $x=32$, giving $4n=32\Rightarrow n=8.$ This is valid.

So A is possible.

Option B: 1088

We need $x(x+1)=1088.$ But $32\cdot 33=1056,\qquad 33\cdot 34=1122.$ Since $1088$ lies between these two consecutive products, no integer $x$ satisfies it.

So B is not possible.

Option C: 1120

We need $x(x+1)=1120.$ But $33\cdot 34=1122,$ and no consecutive integers multiply to $1120$.

So C is not possible.

Option D: 1332

We need $x(x+1)=1332.$ Now $36\cdot 37=1332,$ so $x=36$, hence $4n=36\Rightarrow n=9.$ This is valid.

So D is possible.

6. Therefore, the possible values are $\boxed{1056 \text{ and } 1332}.$ So the correct options are $\boxed{A, D}.$