In vacuum (and approximately in air) the field of view has angular width in radians
theta[air]=2 arctan(L/2 f)
where f is the focal length and L is the width of the sensor, and theta is in radians. But, usually field of view is quoted for the diagonal, and in that case L=sqrt(w^2+h^2), where w is width and h is height.
If the lens is directed through a flat port into water, then the field of view has angular width given by Snell's Law:
theta[water] = 2 arcsin( sin( theta[air] /2)/1.333 )
where 1.333 is the index of refraction of water.
The D70 sensor is 23.7 by 15.6 mm, so it measures 28.37mm diagonally. Then, an 18mm lens has field of view in air of
theta[D70,air] =2 arctan(28.37/2*18) = 1.335 radians
Convert radians to degrees by multiplying by (180/pi) = 57.29:
theta[D70,air] = 1.335 * 57.29 = 76 degrees :: is the field of view of a D70 with 18mm lens in air.
A full-frame sensor is 36x24 mm, so 43.27mm diagonally. Then, a 27mm lens would have field of view
theta[FF,air] =1.34 radians = 77 degrees :: is the field of view of a FF sensor with 18mm lens in air.
This is almost the same as the 18mm lens for the D70. The ratio. (27 mm/18 mm) is the "Nikon factor" of 1.5, to account for the smaller sensor.
In water, the D70 with a flat port would have field of view
theta[D70,water] = 2 arcsin( sin( 1.335 /2) /1.333) = 0.965 radians = 55 deg :: FOV of a D70 w/ 18mm lens in water.
With a little algebra, you can show that this is the same field of view as a 41mm lens with a full-frame sensor in air, or with your D70 in air with a 27mm lens.
I hope this helps!