How to prove that EPR outcomes have equal probability no matter the basis?

Let $rho_{AB}$ be a quantum state shared between two parties, Alice and Bob. Suppose Alice performs a POVM measurement ${M_i}_i$ on her half of the state. Then the probability that Alice obtains outcome $i$ is given by the Born rule as $$ p(i) = mathrm{Tr}[rho_{AB}(M_i otimes I)]. $$ But whenever we have the trace of a multipartite operator we can always perform a partial trace first, i.e., $mathrm{Tr}[X] = mathrm{Tr}[mathrm{Tr}_B[X]]$. So in this case $$ p(i) = mathrm{Tr}[mathrm{Tr}_B[rho_{AB}(M_i otimes I)]] = mathrm{Tr}[rho_{A} M_i] $$ where $rho_A = mathrm{Tr}_B[rho_{AB}]$. For the EPR states, if you calculate $rho_A$ (you should verify this) you'll find it is equal to $I/2$ (the maximally mixed state). Thus for the EPR states $$ p(i) = mathrm{Tr}[I/2 M_i] = frac12 mathrm{Tr}[M_i]. $$ Now if $M_i$ is a rank one qubit projector (which I'm assuming is what you meant by a measurement) then $mathrm{Tr}[M_i] = 1$.

The local state (described by density matrix) of each qubit in EPR state is begin{equation} rho=frac{1}{2}begin{pmatrix} 1 & 0 0 & 1 end{pmatrix} end{equation} It does not depend on basis, so both outcomes have 50% probabilities in every measurement basis.

Proving directly the expression for qubit's local state is a little long.

The EPR state is pure 2-qubit state $$ |Psirangle_{EPR}= frac{1}{sqrt{2}}(|0rangle|0rangle+|1rangle|1rangle) $$ or in the standard basis $$ |Psirangle_{EPR}= frac{1}{sqrt{2}}begin{pmatrix}{11} end{pmatrix} $$

The 2-qubit density matrix is $$ rho_{AB}=|Psirangle_{EPR} langle Psi|_{EPR}= frac{1}{sqrt{2}}begin{pmatrix}11 end{pmatrix} cdot frac{1}{sqrt{2}}begin{pmatrix}1 & 0 & 0 & 1 end{pmatrix}=frac{1}{2}begin{pmatrix} 1&0&0&1 0&0&0&0 0&0&0&0 1&0&0&1 end{pmatrix} $$ and the local state of say qubit $A$ is given by the partial trace over qubit $B$ $$ rho_A=tr_B[rho_{AB}]=frac{1}{2}begin{pmatrix} 1&0 0&1 end{pmatrix} $$ The same expression is obtained for qubit $B$