The method that you propose is certainly the most straightforward, but it is a bit tedious. You can speed things up by breaking it up into four sums and dealing with each of them separately:
$$\sum_{i=-1}^5\sum_{j=-5}^11+2\sum_{i=-1}^5\sum_{j=-5}^1i+3\sum_{i=-1}^5\sum_{j=-5}^1j+4\sum_{i=-1}^5\sum_{j=-5}^1ij\;.$$
The first one is easy: you add once for each possible pair $\langle i,j\rangle$, and there are $7^2=49$ possible pairs, so the total is $49$. (There are $7$ values of $i$ and $7$ values of $j$.)
The second one takes just a little more work. For each value of $i$ in the outer sum you’re simply adding up $7$ copies of $i$ in the inner sum, so
$$\sum_{i=-1}^5\sum_{j=-5}^1i=7\sum_{i=-1}^5i\;,$$
and that last summation is just an arithmetic series.
For the third one you can sum the arithmetic series $\sum_{j=-1}^1j$ and then observe that the outer sum is just adding up $7$ copies of this total.
That leaves the most interesting one. Notice that $i$ is a constant in the inner summation: it doesn’t depend on $j$. Thus,
$$\sum_{i=-1}^5\sum_{j=-5}^1ij=\sum_{i=-1}^5i\sum_{j=-5}^1j\;.$$
The inner summation is now just the same arithmetic series as in the third summation, so you can evaluate it and pull out the sum as a constant multiplier. All that remains then is to sum the arithmetic series $\sum_{i=-1}^5i$ and multiply the result by that multiplier (and by $4$, of course).