The short answer is, make the holes as large as you can and cover them with a screen and/or hide them in the background (i.e. create an overhang with the hole under/behind). Or, make more smaller holes. Given what I've calculated below in the long answer using reasonable assumptions for the amount of flow you might have going through the intakes to filter a 125g, I'd say you're probably going to want to make your intake holes through the background larger than 1 inch. You're probably going to want at least 2 inch holes. Keep reading for the long answer and the math that you can use to calculate the hole size. Or at least post your filter flow rate and I can give you a more informed short answer.

The long answer depends on how much flow needs to go through the hole(s) (and fair warning, the long answer is long and involves math/algebra). Then you'd want to decide how much you can allow the water level in the chamber behind the hole to drop. In order to flow through the hole, a pressure differential needs to be created and the way that is created is by having a different water level on either side of the hole. So your intake pulls water out of the chamber behind the hole, causing the water level to drop. When the water level drops, water is pulled from the tank through the hole into the chamber. The water level in the chamber will drop until a pressure differential is generated large enough to cause the flow through the hole to equal the flow through the intake. The larger the hole, the less the water level needs to drop. If the hole is large enough, the drop in water level will be insignificant.

What you have is the engineering equivalent of a submerged orifice:

The equation at the bottom is what you want to use.

- Q is the flow you need to pass through the hole. With a filter you usually know Q in gph. This needs to be converted to in^3/s for this equation to work. [gph]*231[in^3/gal]/60[min/hour]/60[sec/min] = [in^3/s][/*]
- C is what's referred to as the discharge coefficient and is usually .6 to .7[/*]
- a is the area of the hole. the area of a circular hole is pi*r^2. so a 1 inch diameter hole has an area of pi*.5^2 = .7854 in^2[/*]
- g is the gravitational constant which is 386 in/s^2[/*]
- delta h is then how much the water level will drop in the chamber behind the hole.[/*]

delta h is your unknown and a is what you can control, your variable. The rest of the terms are constant. You can rearrange the equation to solve for delta h like below. Or, if you know what you want delta h to be, you can rearrange to solve for what your hole area a needs to be.

idk what your filter flow is but for a 125 let's say you've got 1000gph split between the 2 intakes so 500gph flowing through the 1 inch hole. Doing the unit conversions and math, the water level in the chamber behind the hole would drop 6 inches lower than the water level in the main part of the tank (using .6 for C). So that would tell you how low the filter intake needs to be or how far to submerge the heater.

Delta h is proportional to the square of the flow rate so if your flow is more than what I've assumed, delta h will increase quickly. If your flow is twice what I've assumed for example, delta h will be 4x larger (24inches). You can see that a 1inch hole quickly becomes inadequate. If flow is half what I've assumed, h will be 1/4x as large (1.5inches).

The good new is that delta h is also inversely proportional to the square of the hole area. And, hole area is proportional to the square of the hole diameter. So if you made your hole 2 inches in diameter instead of 1 inch, the area of the hole would be 4x larger, and the delta h would be inversely squared that, or 1/16x as large. So for the same 500gph flowing through a 2inch hole instead of a 1 inch hole, the delta h would only be 3/8 inch rather than 6 inches. Then doing the same sensitivity study on flow, if flow was 1000gph instead of 500gph, delta h would be 1.5 inches. etc.

Another consideration: if you wanted to stick with 1 inch holes only, you could simply add more holes to achieve the same effect on delta h as a single 2 inch hole. But remember when we went from a 1 inch hole to a 2 inch hole, the hole area got 4x larger. So intuitively, you'd need four 1 inch holes to have the same effect as a single 2 inch hole.

If you're comfortable with doing the math, you can use the equations I gave you to size your intake holes. if you know what you want to limit delta h to, you can use the 2nd equation and solve for a instead, and then figure out what size and number of holes gives you that a. As an optimization problem, the goal may be to minimize delta h. since all you really have control over is size and number of intake holes, minimizing delta h means maximizing a. This circles back to the simple answer I gave at the top: make the holes as big as possible.

If you don't want to do the math yourself, just post your filter flow rate and how far you'd be ok with the water dropping in the intake chamber behind the hole and I can run the numbers.