What is the mathematical relationship between Specific Excess Power and an aircraft's acceleration at a given point on an E-M diagram?
You forgot that you accelerate not from standstill, but a considerable speed. So that equivalent potential energy change comes on top of the existing kinetic energy and makes the energy change look much smaller (resulting from the square of speed in kinetic energy). The correct calculation is
$$\frac{\delta E_{pot}}{\delta t} = 800 \;\text{ft/sec}\cdot g=(v_1^2-v_0^2)\cdot \frac{1}{2\cdot \delta t}$$
Your acceleration will decrease as speed increases. If I assume a speed of 800 ft/sec, the resulting acceleration is to a speed of
$$v_1 = \sqrt{v_0^2+2\cdot g\cdot h} = 831.55\;\text{ft/sec}$$
within one second which is 31.55 ft/sec² or just shy of 1 g. The initial acceleration is 1 g and decreases as the aircraft picks up speed. In mathematical terms:
$$\ddot x = SEP\cdot\frac{g}{v} = v_z\cdot\frac{g}{v}$$
When climb speed (or specific excess power) equals flight speed, both cancel, leaving gravitational acceleration equal to horizontal acceleration if all climb potential is used for increasing speed.