The answer below assumes that the payoff provided by the OP is correct. The question appears simple, but still a lot of subtleties to deal with.
We assume that, under the risk-neutral probability measure,
\begin{align*}
dS_i(t) = S_i(t) \big(r dt + \sigma_i dW_i(t)\big),
\end{align*}
for $i=1, \ldots, n$, where $r$ is the interest rate, $\sigma_i$ is the volatility and $\{W_i(t), t\ge 0\}$ is a standard Brownian motion such that $d\langle W_i, W_j\rangle(t) = \rho_{i, j}dt$ for $i\ne j$. Then,
\begin{align*}
\Big(\Pi_{i=1}^nS_i(T)\Big)^{\frac{1}{n}} = \Big(\Pi_{i=1}^nS_i(0)\Big)^{\frac{1}{n}}\exp\bigg(\Big(r -\frac{1}{2n}\sum_{i=1}^n\sigma_i^2\Big)T + \frac{1}{n}\sum_{i=1}^n \sigma_iW_i(T)\bigg).
\end{align*}
Let
\begin{align*}
\sigma = \frac{1}{n}\sqrt{\sum_{i=1}^n \sigma_i^2 + 2\sum_{i\ne j} \rho_{i, j} \sigma_i \sigma_j},
\end{align*}
and
\begin{align*}
\xi = \frac{\frac{1}{n}\sum_{i=1}^n \sigma_iW_i(T)}{\sigma \sqrt{T}}.
\end{align*}
Then, $\xi$ is standard normal, that is, $\xi\sim N(0, 1)$. Let
\begin{align*}
F = \Big(\Pi_{i=1}^nS_i(0)\Big)^{\frac{1}{n}}\exp\bigg(\Big(r -\frac{1}{2n}\sum_{i=1}^n\sigma_i^2 + \frac{1}{2}\sigma^2\Big)T\bigg).
\end{align*}
Then,
\begin{align*}
\Big(\Pi_{i=1}^nS_i(T)\Big)^{\frac{1}{n}} = F e^{-\frac{1}{2}\sigma^2 T + \sigma \sqrt{T} \xi}.
\end{align*}
Moreover, as the derivation of the Black-Scholes formula,
\begin{align*}
E\left(\max\bigg(\Big(\Pi_{i=1}^nS_i(T)\Big)^{\frac{1}{n}} -K, 0\bigg) \right) &=E\left(\max\bigg(F e^{-\frac{1}{2}\sigma^2 T + \sigma \sqrt{T} \xi} -K, 0\bigg) \right)\\
&=F\Phi(d_1) - K \Phi(d_2),
\end{align*}
where $\Phi$ is the cumulative distribution function of a standard normal random variable,
\begin{align*}
d_1 = \frac{\ln \frac{F}{K}+\frac{1}{2}\sigma^2 T}{\sigma \sqrt{T}},
\end{align*}
and
\begin{align*}
d_2 = d_1 - \sigma \sqrt{T}.
\end{align*}
The option value is then given by
\begin{align*}
e^{-rT} \big[F\Phi(d_1) - K \Phi(d_2) \big].
\end{align*}
