The question is basically the title but given a unitary operator $U$ and a computational basis, can we say anything about the complex number below?
$$c = \sum_{i}\langle i \vert U \vert j\rangle$$
I expected that it would be $|c| = 1$ but this does not seem to generally hold.
Welcome to the quantum computing stack exchange.
If you view $\sum_{i} \langle i |$ as a (non-properly normalized) (bra) state $\sqrt{d}\langle \psi |$, where $|\psi\rangle = \frac{1}{\sqrt{d}}\sum_{x \in \{0,1\}^{n}}|x\rangle$, the quantity $c$ becomes just the inner product of $\sqrt{d}\langle \psi |U|j\rangle$. Here, $d =2^{n}$ is the proper normalization constant.
Without imposing any other constraints on $U$, $U|j\rangle$ just becomes some random state $|\phi\rangle$. Thus, $c = \sqrt{d}\langle \psi|\phi\rangle$ and the only thing we can conclude is:
$$ 0 \leq |c| \leq \sqrt{d} $$
You know that the columns of a unitary matrix are orthonormal: https://math.stackexchange.com/questions/1688950/why-do-the-columns-of-a-unitary-matrix-form-an-orthonormal-basis. This means that the squares of the elements in each column should add up to 1, but that doesn't mean the sum of the elements in each column (what you have above) will add up to a value of magnitude 1.
For example, you could have $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},$$ which is clearly unitary.
The squared elements of the first half column add up to $\frac{1}{2}+\frac{1}{2}=1$ as do the squared elements of the second column. However, the elements of the first column itself add up to $\frac{1}{\sqrt{2}}+ \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}}$ while the elements of the second column add up to $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0.$
