Here's an answer to my own question (thanks to Matt Hedden for the approach):

It will be helpful to fix a presentation of $\pi_{1}(Y)$. Let $T_{1},\dots,T_{n}$ be regular neighborhoods of the exceptional fibers. We can write $Y'=Y\setminus(\cup_{j}T_{j})$ as a circle bundle over $S^{2}\setminus(\cup_{i}D_{i})$ with Euler class $b$. Let $\ell\in\pi_{1}(Y')$ correspond to the singular fiber, and for $j=1,\dots, n$ let $m_{j}\in\pi_{1}(Y')$ correspond to the meridians around the deleted disks, oriented so that $m_{j}=\partial D_{j}$. With respect to these generators we can write

$$\pi_{1}(Y')=\langle m_{j},\ell\rangle/([m_{j},\ell],m_{1}\cdots m_{n}\ell^{-b}).$$

After Dehn filling to obtain $Y$, we obtain the presentation

$$\pi_{1}(Y)=\langle m_{j},\ell\rangle/([m_{j},\ell],m_{1}\cdots m_{n}\ell^{-b},m_{j}^{\alpha_{j}}\ell^{\beta_{j}}).$$

The meridians $\mu_{j}$ and longitudes $\lambda_{T_{j}}$ of the filling fibered solid tori $T_{j}$ are given by $\mu_{j}=\alpha_{j}m_{j}+\beta_{j}\ell$ and $\lambda_{T_{j}}=-p_{j}m_{j}+q_{j}\ell$, where $p_{j},q_{j}$ are the unique integers which satisfy $q_{j}\alpha_{j}+p_{j}\beta_{j}=1$ and $1\le p_{j}<\alpha_{j}$.

Now fix $1\le i\le n$ and write the Seifert longitude of the exceptional fiber $K_{i}$ as

$$\lambda_{S_{i}}=\lambda_{T_{i}}+N_{i}\mu_{i}\in H_{1}(\partial T_{i}).$$

The Seifert longitude is characterized by the property that it maps to zero under the inclusion $H_{1}(\partial T_{i})\hookrightarrow H_{1}(Y\setminus T_{i})$. The above fundamental group presentation for $Y$ induces a natural presentation

$$H_{1}(Y\setminus T_{i})=\mathbb{Z}\langle m_{j},\ell\rangle/(\alpha_{j}m_{j}+\beta_{j}\ell=0\text{ for all }j\neq i, \sum_{j}m_{j}=b\ell).$$

Solving for $\lambda_{S_{i}}=0$, we see that

$$0=\lambda_{S_{i}}=\lambda_{T_{i}}+N_{i}\mu_{i}=(N_{i}\alpha_{i}-p_{i})m_{i}+(N_{i}\beta_{i}+q_{i})\ell$$ $$=(N_{i}\alpha_{i}-p_{i})(b\ell-\sum_{j\neq i}m_{j})+(N_{i}\beta_{i}+q_{i})\ell$$ $$=(N_{i}\alpha_{i}-p_{i})(b\ell+\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}\ell)+(N_{i}\beta_{i}+q_{i})\ell$$ $$\implies (N_{i}\alpha_{i}-p_{i})(b+\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}})+(N_{i}\beta_{i}+q_{i})=0.$$

Rearranging, we obtain

$$N_{i}=\frac{-q_{i}+p_{i}b+p_{i}\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}}{\beta_{i}+\alpha_{i}b+\alpha_{i}\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}}.$$

Finally using the relation $\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}=-\Big(\frac{\beta_{i}}{\alpha_{i}}+b+\frac{1}{\alpha_{1}\cdots\alpha_{n}}\Big)$ and rearranging, we obtain the explicit formula

$$N_{i}=\frac{\alpha_{1}\cdots\alpha_{n}}{\alpha_{i}}(q_{i}-bp_{i})-p_{i}\sum_{j\neq i}\frac{\alpha_{1}\cdots\alpha_{n}}{\alpha_{i}\alpha_{j}}\beta_{j}.$$

Here's an answer to my own question (thanks to Matt Hedden for the approach):

It will be helpful to fix a presentation of $\pi_{1}(Y)$. Let $T_{1},\dots,T_{n}$ be regular neighborhoods of the exceptional fibers. We can write $Y'=Y\setminus(\cup_{j}T_{j})$ as a circle bundle over $S^{2}\setminus(\cup_{i}D_{i})$ with Euler class $b$. Let $\ell\in\pi_{1}(Y')$ correspond to the singular fiber, and for $j=1,\dots, n$ let $m_{j}\in\pi_{1}(Y')$ correspond to the meridians around the deleted disks, oriented so that $m_{j}=\partial D_{j}$. With respect to these generators we can write

$$\pi_{1}(Y')=\langle m_{j},\ell\;|\;[m_{j},\ell]=m_{1}\cdots m_{n}\ell^{-b}=1\rangle.$$

After Dehn filling to obtain $Y$, we obtain the presentation

$$\pi_{1}(Y)=\langle m_{j},\ell\;|\;[m_{j},\ell]=m_{1}\cdots m_{n}\ell^{-b}=m_{j}^{\alpha_{j}}\ell^{\beta_{j}}=1\rangle.$$

The meridians $\mu_{j}$ and longitudes $\lambda_{T_{j}}$ of the filling fibered solid tori $T_{j}$ are given by $\mu_{j}=\alpha_{j}m_{j}+\beta_{j}\ell$ and $\lambda_{T_{j}}=-p_{j}m_{j}+q_{j}\ell$, where $p_{j},q_{j}$ are the unique integers which satisfy $q_{j}\alpha_{j}+p_{j}\beta_{j}=1$ and $1\le p_{j}<\alpha_{j}$.

Now fix $1\le i\le n$ and write the Seifert longitude of the exceptional fiber $K_{i}$ as

$$\lambda_{S_{i}}=\lambda_{T_{i}}+N_{i}\mu_{i}\in H_{1}(\partial T_{i}).$$

The Seifert longitude is characterized by the property that it maps to zero under the inclusion $H_{1}(\partial T_{i})\hookrightarrow H_{1}(Y\setminus T_{i})$. The above fundamental group presentation for $Y$ induces a natural presentation

$$H_{1}(Y\setminus T_{i})=\mathbb{Z}\langle m_{j},\ell\rangle/(\alpha_{j}m_{j}+\beta_{j}\ell=0\text{ for all }j\neq i, \sum_{j}m_{j}=b\ell).$$

Solving for $\lambda_{S_{i}}=0$, we see that

$$0=\lambda_{S_{i}}=\lambda_{T_{i}}+N_{i}\mu_{i}=(N_{i}\alpha_{i}-p_{i})m_{i}+(N_{i}\beta_{i}+q_{i})\ell$$ $$=(N_{i}\alpha_{i}-p_{i})(b\ell-\sum_{j\neq i}m_{j})+(N_{i}\beta_{i}+q_{i})\ell$$ $$=(N_{i}\alpha_{i}-p_{i})(b\ell+\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}\ell)+(N_{i}\beta_{i}+q_{i})\ell$$ $$\implies (N_{i}\alpha_{i}-p_{i})(b+\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}})+(N_{i}\beta_{i}+q_{i})=0.$$

Rearranging, we obtain

$$N_{i}=\frac{-q_{i}+p_{i}b+p_{i}\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}}{\beta_{i}+\alpha_{i}b+\alpha_{i}\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}}.$$

Finally using the relation $\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}=-\Big(\frac{\beta_{i}}{\alpha_{i}}+b+\frac{1}{\alpha_{1}\cdots\alpha_{n}}\Big)$ and rearranging, we obtain the explicit formula

$$N_{i}=\frac{\alpha_{1}\cdots\alpha_{n}}{\alpha_{i}}(q_{i}-bp_{i})-p_{i}\sum_{j\neq i}\frac{\alpha_{1}\cdots\alpha_{n}}{\alpha_{i}\alpha_{j}}\beta_{j}.$$