By completing the squares of sums, the given equation can be written as
$$\left(x + \frac{1}{2}\right)^2 + \left(y + \frac{3}{2}\right)^2 + \left(z + \frac{5}{2}\right)^2 + \left(w + \frac{7}{2}\right)^2 = 25.$$
By the inequality of arithmetic and quadratic means, we have
$$\frac{\left(x + \frac{1}{2}\right) + \left(y + \frac{3}{2}\right) + \left(z + \frac{5}{2}\right) + \left(w + \frac{7}{2}\right)}{4} \le \sqrt{\frac{\left(x + \frac{1}{2}\right)^2 + \left(y + \frac{3}{2}\right)^2 + \left(z + \frac{5}{2}\right)^2 + \left(w + \frac{7}{2}\right)^2}{4}}$$
We can conclude that $(x + \frac{1}{2}) + (y + \frac{3}{2}) + (z + \frac{5}{2}) + (w + \frac{7}{2}) \le 10$, i.e.
$$x + y + z + w \le 2.$$
Note that the value 2 can indeed be attained, e.g. for $x = 2$, $y = 1$, $z = 0$, $w = -1$.
Therefore, the largest possible value of $x + y + z + w$ is 2.