Finding distances accurate to a few mm is not going to work with LEDs and photodiodes. Light takes about 1 ns per foot. To measure distance with time of arrival you would need resolution of just a few ps. Not only do ordinary LEDs and photodiodes not react that fast, but the circuit to interpret those signals would need to be very very special even if they could.
You may be able to get what you want by using ultrasound instead of light. At about 3 ms per meter, you have some chance of getting reasonable resolution. Typical cheap ultrasound transducers are resonant around 40 kHz. Those will probably not be good enough because the limited bandwidth will get in the way of the time resolution you need. You should look at higher frequency ultrasound transmitters and receivers that are also more broad spectrum. You might have a chance with those.
Otherwise, there are other techniques than time of arrival to determine position. You could still use light if you can determine the direction of the light instead of trying to measure tiny propagation time differences. Two cameras that look for the location of the beacon in theory would be good enough.
The bandwidth you have calculated is not considering the harmonics of the pulse. Consider that the pulse is 14ns long with inbuilt 2ns rise and fall repeating every 24 ns i.e. a symmetrical waveform. That has a fundamental of about 41 MHz but if you are interested in accuracy of the pulse shape then you should go for an op-amp with ten times this bandwidth.
As for slew rate, if your biggest output signal is (say) 2V p-p then it has to be able change 2V in 2ns = 1000 V/us. Now, onto the main show-stopper.
A 1,000,000:1 range of input currents is really optimistic for a conventional TIA because noise will be a problem for low signals (50 nA) but there are some tricks you can do to improve the basic design such as bootstrapping: -
Note the JFET addition to the standard TIA circuit - in effect it reduces the self-capacitance of the photodiode and solves (to a large extent) the inherent noise problem due to photodiode self-capacitance (see this article for more details and this SE answer also).
However, the 50mA Imax requirement is unfeasible (maybe you mean 50 uA). The 50 mA has to be sunk/sourced by the op-amp's output and this will significantly affect the choice of available devices. I doubt that you will find an appropriate device but maybe you have a cunning plan?
Consider also what feedback resistor value you'll need to produce a change in output of (say) 2V. 2V/0.05A = 40 ohms. What will be the output amplitude when 50 nA is applied? I calculate 2 uV and that signal is totally embedded in noise. Why will it be embedded in noise - consider the input noise spec for the op-amp - maybe it's 3 nV / \$\sqrt{Hz}\$. A TIA having a bandwidth of 400 MHz will produce 60 uV RMS of noise at the output. Do the math - the spread of your limits cannot be met with anything I know of.
If you do want to detect 50 nA you will want the feedback resistor to be much higher than 40 ohms. 40 ohms produces an output of 2 uV in a noise of 60 x 6.6 uVp-p. The 6.6 (sigma) number converts RMS gaussian noise to p-p noise with a confidence factor of 99.9%. So the output noise will be 400 uVp-p and, because of this, you will want your smallest signal to be (say) ten times this value at 4 mVp-p.
A 4mV output change from a 50nA input change implies a feedback resistor value of 80 kohm (not 40 ohm). But this raises the problem of the photodiode capacitance - at 100 MHz a 12pF cap has an impedance of 133 ohms. This capacitance and the 80k feedback resistor amplify the op-amps internal noise by a factor of ~600. In other words you cannot asssume that the photodiode's self capacitance is negligible. It is the "total player" in defining the noise you will get. My previous "simplified" analysis assumed the op-amp is unity gain but it isn't. The 3dB point is \$\frac{1}{2\pi RC}\$ where R = 80k and C = 12pF i.e. F = 165 kHz. In other words from 165 kHz upwards the noise gain of the op-amp rises by 6 dB per octave and at 100 MHz the noise gain (as previously mentioned) is 600 and your TIA idea is toast!
There are some mitigations; the op-amp doesn't have infinite bandwidth so although the noise gain rises from 165 kHz it won't hit a peak of 600 at 100 MHz because if it did your op-amp would have a gain-bandwidth product of 60 GHz and they haven't been made yet.
So, if you say that a decent op-amp has a GBWP of 1 GHz, at 100 MHz the noise gain can never be more than ten. At 10 MHz the noise gain could be 100 but this is determined by the 12 pF and 80 kohm feedback resistor i.e. noise gain will actually be: -
\$1 + \dfrac{80,000}{X_C}\$ = 61
This is something like what your noise gain is going to look like spectrally: -
\$C_{sh}\$ is the photodiodes shunt capacitance i.e. the 12 pF. \$C_{i}\$ is the input capacitance of the op-amp (not negligible of course and adds to the problem). \$R_{f}\$ and \$C_{f}\$ is the feedback resistor and any capacitance in parallel with it (including parasites).
Somewhere in the middle of the raised portion of the graph is 10 MHz. Maybe consider 20 MHz as a spot frequency for examination; open loop gain might be 50 so this is going to be the noise gain limit. What about 5 MHz - the 1+R/Xc ratio will be about 31 and this is going to be the limit i.e. no longer dependent on the open loop gain.
It looks to me like the noise gain is peaking about 10 MHz and is still going to be a big problem for a 50nA input signal and a 80 kohm feedback resistor.
The trouble is, you have to increase the feedback resistor to bring your signal out from the noise and by the looks of it by something like ten times and, this causes another problem. 800 kohm and maybe 1pF of parasitic capacitance in parallel gives a cut-off frequency of 198 kHz i.e. above this the signal drops at 6 dB per decade i.e. it's a low pass filter. You were aiming for a bandwidth of maybe 400 MHz and you are crippled at anything above 200 kHz. Even if the feedback resistor were 80 kohm the BW would still only be 2 MHz.
Summary - alter your spec or hugely adjust your expectations.
It's also worth spending an hour or two using ADI's photodiode design wizard. Even if you don't use one of their devices you can learn a lot by experimenting with it.
Best Answer
1 - Noise pickup on the leads is not what you need to worry about. If nothing else, you can put your detector in a shielded enclosure, with filtered power in and a hole just large enough for the IR to illuminate the array.
2 - What you do need to worry about is (as Kevin White commented) the effects of those long traces on the input capacitance of the op amp. Unless you properly compensate with a capacitor across the feedback resistor of your TIA, the circuit will oscillate. Fortunately, your frequency requirements (<100 kHz) are fairly modest, and I'd guess you should be able to get a decent signal amplitude.
3 - You may need to go to a multi-stage amplifier chain, with the maximum gain of the TIA (set by the feedback resistor) being limited by the feedback capacitor. Subsequent amplifiers will degrade the SNR of the signal slightly, but that's unavoidable.
4 - 40 LTC6268s in close proximity is, frankly, a disaster waiting to happen unless you are very careful about ground paths and power decoupling. Whenever you have multiple high-frequency op amps you need to take care that a signal on one does not affect a neighbor due to coupling through the power supply lines. And, at 100 MHz open-loop bandwidth, the LTC 6268 is a good example of a high-frequency op amp. You might be well-advised to go with an op amp with a lower bandwidth, such as 10 MHz.
5 - Given the fact that your photodiode array has a reverse bias limit of 6 volts, the example TIA shown on page 1 of the 6282 data sheet is an excellent place to start, keeping in mind the need for an overt feedback cap. 1 uA / 20k will give a signal of 20 mV, and you may (or may not) be able to get away with a larger feedback resistor. You can then boost the signal with a subsequent op amp. Since your bandwidth isn't too great, you should be able to provide enough gain without too much extra noise being introduced.