No, this doesn't work for $n \geq 2$. Roughly, I will show that, for almost any $x_{ij}$, there are $P_i$ with $P_i(x_{ij})=P_j(x_{ij})$. So the $\binom{n+3}{2}-1$ values $x_{ij}$ for $\{ i,j \} \neq \{ 1,2 \}$ don't determine the last one. In order to state this precisely, I need to be careful about degenerate solutions.
If $P$ is any polynomial at all, and I take $P_1=P_2 = \cdots = P_{n+3} = P$, we will of course have $P_i(x) = P_j(x)$ for any $x$. Let's call this a "fully degenerate solution". It also seems natural to exclude solutions where any pair of the $P_i$ are equal. Let's say that a solution is "slightly degenerate" if $P_i = P_j$ for some $i \neq j$. Fix $n \geq 2$. What I will actually show is that, for any $x_{ij}$ at all, there are solutions which are not fully degenerate and, for generic $x_{ij}$, there are solutions which are not even slightly degenerate.
To exclude fully degenerate solutions, we just make a dimension count. The vector space of $(n+3)$-tuples of degree $n$ polynomials has dimension $(n+3)(n+1)$. Imposing $P_i(x_{ij}) = P_j(x_{ij})$ is $\binom{n+3}{2}$ linear conditions. So the space of solutions has dimension $\geq (n+3)(n+1) - \binom{n+3}{2} = \tfrac{n(n+3)}{2}$. The space of fully degenerate solutions is a vector space of dimension $n+1$. For $n \geq 2$, we have $\tfrac{n(n+3)}{2} > n+1$, so there is a solution which is not fully degenerate.
We now prove the claim about slightly degenerate solutions. Let $X$ be the $\binom{n+3}{2}$ dimensional space of possible $x_{ij}$'s. Let $U_{ij} \subset X$ be those $(x_{ij})$ for which there is a solution with $P_i \neq P_j$. Note that we have $X = \bigcup_{i<j} U_{ij}$, and that all the $U_{ij}$ are isomorphic to each other by permuting coordinates.
Let $\overline{U}_{ij}$ be the closure of $U_{ij}$ in the Zariski topology. I claim that $\overline{U}_{ij} = X$ for each $(i,j)$. First of all, since all the $U_{ij}$ are isomorphic to each other by permuting coordinates, they either all have closure $X$ or none of them do. If all of their closures are smaller, then $\dim \overline{U}_{ij} < \dim X$, so we can't have $X = \bigcup_{i<j} U_{ij}$, a contradiction. So $U_{ij}$ is dense in $X$.
Now, $U_{ij}$ is constructible in the Zariski topology, so it contains a dense open subset $V_{ij}$. But then $V:=\bigcap_{i<j} V_{ij}$ is a dense open subset of $X$. By definition, if $(x_{ij}) \in V$ then, for each $(i,j)$, there is a solution with $P_i \neq P_j$. But then the generic solution for such an $(x_{ij})$ must have all the $P_i$ distinct.
Remark When $n=1$, we have $6$ linear equations in $8$ varables, so we generically DO expect the only solutions to be the 2 dimensional space of degenerate ones. Indeed, one can rederive Menelaus theorem by writing down the $6 \times 8$ matrix and setting one of its $6 \times 6$ minors to $0$.
In the above answer, I copied the OP's $n+3$ without thinking about it too much. If we switch to $2n+2$ polynomials instead, there will be a positive result. We can normalize one of the polynomials to be $0$, then the other $2n+1$ polynomials have $(2n+1)(n+1)$ coefficients between them. Imposing equality at all but $x_{12}$ is $\binom{2n+2}{2}-1 = (2n+1)(n+1)-1$ linear conditions. So, generically, there will be a unique solution. Then $x_{12}$ will be determined as one of the $n$ roots of $P_1(x) - P_2(x)$.
