I'm not sure why you emphasize linear combinations with coefficients $\pm 1$ to the exclusion of other possible rational coefficients.  

In any case, the degree of $\mathbf Q(\sqrt{a_1},\ldots,\sqrt{a_n})$ over $\mathbf Q$ equals the order of $\langle a_1,\ldots,a_n\rangle$ in $\mathbf Q^\times/(\mathbf Q^\times)^2$, which is basically counting how multiplicatively independent the $a_i$'s are modulo rational squares. An approach to proving multiplicative independence of $a_i$'s mod squares implies their square roots are linearly independent over $\mathbf Q$ by using reduction mod $p$ for a large prime $p$ can be read [here][1].


[1]:https://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/