Note that, as in this question, for $s\ge t\ge 0$,
\begin{align*}
n_s = e^{-a_n(s-t)}n_t + \int_t^s \theta_n(u)e^{-a_n(s-u)} du + \int_t^s \sigma_n e^{-a_n(s-u)} dW^n_u,
\end{align*}
and
\begin{align*}
r_s = e^{-a_r(s-t)}r_t + \int_t^s (\theta_r(u) -\rho_{r,n}\sigma_n\sigma_r) e^{-a_r(s-u)} du + \int_t^s \sigma_r e^{-a_r(s-u)} dW^r_u.
\end{align*}
Moreover,
\begin{align*}
\int_t^T n_s ds
= \frac{1}{a_n}\Big(1-e^{-a_n(T-t)} \Big) n_t &+ \int_t^T\!\! \frac{\theta_n(u)}{a_n}\Big(1-e^{-a_n(T-u)} \Big)du \\
&+ \int_t^T \!\!\frac{\sigma_n}{a_n}\Big(1-e^{-a_n(T-u)} \Big)dW_u^n,
\end{align*}
and
\begin{align*}
\int_t^T r_s ds=
\frac{1}{a_r}\Big(1-e^{-a_r(T-t)} \Big) n_t &+ \int_t^T\!\! \frac{\theta_r(u)-\rho_{r,n}\sigma_n\sigma_r}{a_r}\Big(1-e^{-a_r(T-u)} \Big)du \\
&+ \int_t^T \!\!\frac{\sigma_r}{a_r}\Big(1-e^{-a_r(T-u)} \Big)dW_u^r.
\end{align*}
Let $B_n(t, T) = \frac{1}{a_n}\Big(1-e^{-a_n(T-u)} \Big)$, and $B_r(t, T) = \frac{1}{a_r}\Big(1-e^{-a_r(T-u)} \Big)$.
Then,
\begin{align*}
\int_t^T n_s ds &= B_n(t, T) n_t + \int_t^T \theta_n(u) B_n(u, T) du + \int_t^T \sigma_n B_n(u, T) dW_u^n,
\end{align*}
and
\begin{align*}
\int_t^T r_s ds &= B_r(t, T) r_t + \int_t^T (\theta_r(u)-\rho_{r,n}\sigma_n\sigma_r) B_r(u, T) du + \int_t^T \sigma_r B_r(u, T) dW_u^r.
\end{align*}
Moreover,
\begin{align*}
E\left(e^{\int_t^T (n_s-r_s) ds} \mid \mathcal{F}_t \right) &=
e^{B_n(t, T) n_t-B_r(t, T) r_t+\int_t^T \theta_n(u) B_n(u, T) du -\int_t^T (\theta_r(u)-\rho_{r,n}\sigma_n\sigma_r) B_r(u, T) du}\\
&\quad \times e^{\frac{1}{2}\int_t^T \sigma_n^2 B_n(u, T)^2 du+\frac{1}{2}\int_t^T \sigma_r^2 B_r(u, T)^2 du - \int_t^T \rho_{r,n}\sigma_n\sigma_r B_n(u, T)B_r(u, T)du}.
\end{align*}
Moreover,
\begin{align*}
\int_t^T \sigma_n^2 B_n(u, T)^2 du &= -\frac{\sigma_n^2}{a_n^2}\big(B_n(t, T) -T+t\big)-\frac{\sigma_n^2}{2a_n}B_n(t, T)^2,\\
\int_t^T \sigma_r^2 B_r(u, T)^2 du &= -\frac{\sigma_r^2}{a_r^2}\big(B_r(t, T) -T+t\big)-\frac{\sigma_r^2}{2a_r}B_r(t, T)^2
\end{align*}
and
\begin{align*}
\int_t^T B_n(u, T)B_r(u, T)du = \frac{1}{a_na_r}\left[T-t - B_n(t, T)-B_r(T, t)+\frac{1}{a_n+a_r}\left(1-e^{-(a_n+a_r)(T-t)}\right) \right].
\end{align*}
Therefore,
\begin{align*}
E\left(e^{\int_t^T (n_s-r_s) ds} \mid \mathcal{F}_t \right)
&= A_n(t, T) A_r(t, T) C(t, T) e^{B_n(t, T) n_t-B_r(t, T) r_t},
\end{align*}
where
\begin{align*}
A_n(t, T) &= e^{\int_t^T \theta_n(u) B_n(u, T) du -\frac{\sigma_n^2}{2a_n^2}\big(B_n(t, T) -T+t\big)-\frac{\sigma_n^2}{4a_n}B_n(t, T)^2}, \\
A_r(t, T) &= e^{-\int_t^T (\theta_r(u)-\rho_{r,n}\sigma_n\sigma_r) B_r(u, T) du -\frac{\sigma_r^2}{2a_r^2}\big(B_r(t, T) -T+t\big)-\frac{\sigma_r^2}{4a_r}B_r(t, T)^2},
\end{align*}
and
\begin{align*}
C(t, T) = e^{\frac{-\rho_{r,n}\sigma_n\sigma_r}{a_na_r}\left[T-t - B_n(t, T)-B_r(T, t)+\frac{1}{a_n+a_r}\left(1-e^{-(a_n+a_r)(T-t)}\right) \right]}
\end{align*}
Note that this is similar to the Hull-White zero-coupon bond pricing formula.