First, when solving problems like this one, you have to make sure that both salts possess good solubility and that no double salt is precipitating:
$$
\begin{align}
\ce{NaNO3 &<=>> Na+ + NO3-}\label{rxn:R1}\tag{R1}\\
\ce{Ca(NO3)2 &<=>> Ca^2+ + 2NO3-}\label{rxn:R2}\tag{R2}
\end{align}
$$
Both nitrates are indeed well soluble in water (equilibrium is shifted to the far right), and no poorly soluble side-product is formed, e.g. we are likely to have a true solution consisting of hydrated cations $\ce{Na+}$, $\ce{Ca^2+}$ as well as $\ce{NO3-}$ anions.
As you already suggested, one can indeed use total amounts of substances $n_i$ and the preservation of volumes $V_i$ to find the final molar concentration on nitrate (here indices $1$ and $2$ refer to initial sodium and calcium nitrate solutions, respectively):
$$c(\ce{NO3-}) = \frac{n_1(\ce{NO3-}) + n_2(\ce{NO3-})}{V_1 + V_2}\label{eqn:1}\tag{1}$$
According to the stoichiometry (see \eqref{rxn:R1} and \eqref{rxn:R2}):
$$n_1(\ce{NO3-}) = n(\ce{NaNO3}) = c(\ce{NaNO3})\cdot V_1\tag{2.1}$$
$$n_2(\ce{NO3-}) = 2\cdot n(\ce{Ca(NO3)2}) = 2\cdot c(\ce{Ca(NO3)2})\cdot V_2\tag{2.2}$$
Finally, we can plug all the known quantities back to \eqref{eqn:1}:
$$
\begin{align}
c(\ce{NO3-}) &= \frac{c(\ce{NaNO3})\cdot V_1 + 2\cdot c(\ce{Ca(NO3)2})\cdot V_2}{V_1 + V_2}\tag{3}\\
&= \frac{\pu{1.23 mol L-1}\cdot\pu{16.3 mL} + 2\cdot\pu{0.823 mol L-1}\cdot\pu{6.28 mL}}{\pu{16.3 mL} + \pu{6.28 mL}}\\
&= \pu{1.35 mol L-1}
\end{align}
$$
