It is not possible even for four points. Consider the following distance table:
$ \sigma = \begin{pmatrix} 0 & 18 & 18 & 10 \\ 18 & 0 & 18 & 10 \\ 18 & 18 & 0 & 10 \\ 10 & 10 & 10 & 0 \end{pmatrix} $
This table satisfies the conditions: it is symmetric and the triangle inequalities hold. However, this table cannot be realized by vectors in $\mathbb{R}^3$. Let's try to construct such vectors $A_1, \dotsc, A_4$. Without loss of generality we can take $A_1 = 0$. Then $||A_k||^2 = \sigma_{1,k}^2$ for $k \in \{2,3,4\}$. For $i, j \geq 2$ and $i \neq j$ the following equality must hold:
$ \sigma_{i,j}^2 = ||A_i - A_j||^2 = ||A_i||^2 + ||A_j||^2 - 2\langle A_i, A_j\rangle = \sigma_{1,i}^2 + \sigma_{1,j}^2 - 2 \langle A_i, A_j \rangle $
and so $\langle A_i, A_j \rangle = (\sigma_{1,i}^2 + \sigma_{1,j}^2 - \sigma_{i,j}^2)/2$. The Gram matrix $A_{i,j} = \langle A_{i+1}, A_{j+1} \rangle$ for the vectors $A_2, A_3, A_4$ must therefore be
$ A = \begin{pmatrix} 324 & 162 & 162 \\ 162 & 324 & 162 \\ 162 & 162 & 100 \end{pmatrix}. $
However, $\det(A) = -629856 < 0$ and this is impossible for a Gram matrix in Euclidean three-space since such a matrix must be positive (semi-)definite.