Let $k_i = m_i \omega_i^2$, and we have
\begin{equation}
H\big(\{p_i\},\{x_i\}\big) = \sum_{i=1}^{N} \frac{p_i^2}{2m_i} + \sum_{i=0}^{N} \frac{k_i}{2}(x_{i+1}-x_i)^2,
\end{equation}
where $x_{{\scriptscriptstyle N+1}} = x_{{\scriptscriptstyle 0}} = 0$. The partition function is given by
\begin{equation}
\begin{split}
Z &= \int \left(\prod_{i=1}^N dp_i\right) \left(\prod_{i=1}^N dx_i\right) \,\exp\left[-\frac{H\big(\{p_i\},\{x_i\}\big)}{k_{B}T}\right]\\
&= \int \left(\prod_{i=1}^N dp_i\right)\, \exp\left(-\frac{1}{k_{B}T}\sum_{i=1}^{N} \frac{p_i^2}{2m_i}\right)\int \left(\prod_{i=1}^N dx_i\right)\, \exp\left[-\frac{1}{k_{B}T}\sum_{i=0}^{N}\frac{k_{i}}{2}(x_{i+1} - x_{i})^2\right]\\
&=I_p I_x\,,
\end{split}
\end{equation}
where $I_p$ and $I_x$ respectively represent the $p$ and $x$ integrals in the second line.
The $p$ integral is a product of Gaussian integrals, which gives
\begin{equation}
I_p = (2\pi k_B T)^{N/2}\sqrt{\prod_{i=1}^{N}m_{i}}
\end{equation}
To evaluate $I_x$, we first make the change of variables $x_{i}^\prime = x_{i} / \sqrt{k_B T} $. Then, $I_x = (k_B T)^{N/2} \tilde{I}_{x}$, where
\begin{equation}
\begin{split}
\tilde{I}_{x} &= \int\left(\prod_{i=1}^N dx_i^\prime\right) \,\exp\left[-\sum_{i=0}^{N}\frac{k_{i}}{2}(x_{i+1}^\prime - x_{i}^\prime)^2\right]\\
&= \int\left(\prod_{i=1}^{N+1} dx_i^\prime\right) \,\delta(x_{N+1}^\prime)\,\exp\left[ -\sum_{i=0}^{N}\frac{k_{i}}{2}(x_{i+1}^\prime - x_{i}^\prime)^2\right].
\end{split}
\end{equation}
In fact, for all practical purposes, it suffices to stop at this point. That is,
\begin{equation}
Z = I_p I_x = (2\pi)^{N/2} (k_B T)^N \tilde{I}_x,
\end{equation}
where $\tilde{I}_x$ is independent of any thermodynamic variable of the system (in this case $T$).
Still, it is possible to evaluate the integral $\tilde{I}_x$. To do this, we make another change of variables $y_{i} = x_{i}^\prime - x_{i - 1}^\prime \ \ (i = 1, 2, \ldots, N)$. The associated Jacobian determinant is equal to 1, and it follows that $x_{N+1}^\prime = \sum_{i=1}^{N+1} y_{i}$. Also, we can represent the delta function as
\begin{equation}
\delta(x_{N+1}^\prime) = \int \frac{dq}{2\pi}\, \exp(iqx_{N+1}^\prime)
\end{equation}
Then,
\begin{equation}
\begin{split}
\tilde{I}_x &= \int\frac{dq}{2\pi}\left(\prod_{i=0}^{N} dy_{i+1}\right) \,\exp\big[iq({\scriptstyle \sum_{i=0}^{N} y_{i+1}})\big] \exp\left( -\sum_{i=0}^{N}\frac{k_{i}}{2}y_{i+1}^2\right)\\
&= \int \frac{dq}{2\pi} \prod_{i=0}^{N}\Bigg[\int dy_{i+1} \exp\bigg(-\frac{k_{i}}{2}y_{i+1}^2 + iqy_{i+1}\bigg)\Bigg]\\
&= \int \frac{dq}{2\pi} \prod_{i=0}^{N} \sqrt{\frac{2\pi}{k_i}}\exp\left(-\frac{q^2}{2k_i}\right)\\
&=\frac{(2\pi)^{N/2}}{\sqrt{\Big(\sum_{i=0}^{N}\frac{1}{k_{i}}\Big)\prod_{i=0}^{N} k_i}}.
\end{split}
\end{equation}
Collecting the results we have derived so far, we obtain
\begin{equation}
Z = I_p I_x = (2\pi k_B T)^{N} \sqrt{\frac{\prod_{i=1}^{N}m_{i}}{\Big(\sum_{i=0}^{N}\frac{1}{k_{i}}\Big)\prod_{i=0}^{N} k_i}}.
\end{equation}
