Yes, the sample variance-covariance matrix is singular if the dimension $p$ is equal to sample size $n$.
Proof:
Let $x_1, .., x_p$ be $p$ vectors of $\mathbb{R}^p$. Let's denote their mean vector by $\bar{x} = \frac{1}{n}\sum_i x_i $ .
Define their sample variance-covariance matrix $$S = \frac{1}{n}\sum_i (x_i - \bar{x}) (x_i - \bar{x}) ^ T = U U^T$$ where $U$ is the (square) matrix whose columns are $x_i - \bar{x}$.
Since $U$ is squared, $$\mathrm{det}(S) = \mathrm{det}(U)^2$$ so if $U$ is singular, $S$ is. Now we can see that $U$ is singular since :
$$U \times
\left( \begin{array}[c]
\;1\\
\vdots\\
1
\end{array}\right) = \left(\begin{array}[c]
1 \sum_i (x_{i,1} - \bar{x}_1)\\
\;\;\;\;\;\;\vdots\\
\sum_i (x_{i,p} - \bar{x}_p)
\end{array}\right) = \left( \begin{array}[c]
\; 0\\
\vdots\\
0
\end{array}\right)$$