Hidden in the comment by Victor Klepstyn is a really nice argument. Since nobody upvoted that comment yet (I just did) it's probably worth expanding it.

From a probabilistic approach it's more natural to show the equivalent statement that the expectation of $(\det A)^2$ equals $\frac{(n+1)!}{(n(n+1))^n}$. The trick is to realize the uniform measure on the simplex as the distribution of $$ (X_1,\dots,X_n) := \frac{(Y_1,\dots,Y_n)}{Y_1 + \cdots +Y_n} $$ where $(Y_i)$ are i.i.d. exponential random variables (say, of parameter 1). All you need to know is that $\mathbf{E} [Y_i] =1$ and $\mathbf{E} [Y_i^2] =2$. Moreover in this representation $(X_i)$ are independent from $S:=Y_1+\cdots+Y_n$. It follows that we have the following identity in distribution $$ B = D A $$ where $B$ is a matrix with i.i.d. exponential entries and $D$ a diagonal matrix whose entries are independent copies of $S$, with $A$ and $D$ moreover independent. Therefore $$ \mathbf{E} [\det B^2] = \mathbf{E} [\det D^2] \cdot \mathbf{E} [\det A^2] .$$ It is very easy to check that $\mathbf{E} [\det D^2] = (n(n+1))^n$, and $\mathbf{E} [\det B^2]$ can be expanded as $$ \sum_{\sigma,\tau \in \mathfrak{S}_n} \varepsilon(\sigma) \varepsilon(\tau) 2^{Fix(\sigma \tau^{-1})} = n! \sum_{\pi \in \mathfrak{S}_n} \varepsilon(\pi) 2^{Fix(\pi)} = n! \det \begin{pmatrix} 2 & 1 & \cdots & 1 \\ 1 & 2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \vdots \\ 1 & \cdots & \cdots & 2 \end{pmatrix} = (n+1)n!$$ (Here $Fix$ denotes the number of fixed points.)

Hidden in the comment by Victor Kleptsyn is a really nice argument. Since nobody upvoted that comment yet (I just did) it's probably worth expanding it.

From a probabilistic approach it's more natural to show the equivalent statement that the expectation of $(\det A)^2$ equals $\frac{(n+1)!}{(n(n+1))^n}$. The trick is to realize the uniform measure on the simplex as the distribution of $$ (X_1,\dots,X_n) := \frac{(Y_1,\dots,Y_n)}{Y_1 + \cdots +Y_n} $$ where $(Y_i)$ are i.i.d. exponential random variables (say, of parameter 1). All you need to know is that $\mathbf{E} [Y_i] =1$ and $\mathbf{E} [Y_i^2] =2$. Moreover in this representation $(X_i)$ are independent from $S:=Y_1+\cdots+Y_n$. It follows that we have the following identity in distribution $$ B = D A $$ where $B$ is a matrix with i.i.d. exponential entries and $D$ a diagonal matrix whose entries are independent copies of $S$, with $A$ and $D$ moreover independent. Therefore $$ \mathbf{E} [\det B^2] = \mathbf{E} [\det D^2] \cdot \mathbf{E} [\det A^2] .$$ It is very easy to check that $\mathbf{E} [\det D^2] = (n(n+1))^n$, and $\mathbf{E} [\det B^2]$ can be expanded as $$ \sum_{\sigma,\tau \in \mathfrak{S}_n} \varepsilon(\sigma) \varepsilon(\tau) 2^{Fix(\sigma \tau^{-1})} = n! \sum_{\pi \in \mathfrak{S}_n} \varepsilon(\pi) 2^{Fix(\pi)} = n! \det \begin{pmatrix} 2 & 1 & \cdots & 1 \\ 1 & 2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \vdots \\ 1 & \cdots & \cdots & 2 \end{pmatrix} = (n+1)!$$ (Here $Fix$ denotes the number of fixed points.)